Step 1: Classification of the Differential Equation

The given equation is $\frac{dy}{dx} + ay = e^{mx}$. This is a first-order linear ordinary differential equation of the form:

$\frac{dy}{dx} + P(x)y = Q(x)$

[Where $P(x)$ and $Q(x)$ are constants or functions of $x$ alone]. Here, $P(x) = a$ and $Q(x) = e^{mx}$.

Step 2: Calculating the Integrating Factor (IF)

The integrating factor is defined as $IF = e^{\int P(x) \, dx}$.

$IF = e^{\int a \, dx} = e^{ax}$

[Per the standard method for solving linear differential equations].

Step 3: Multiplying and Integrating

The general solution for a linear differential equation is given by:

$y \cdot (IF) = \int [Q(x) \cdot (IF)] \, dx + C$

Substituting the values derived in Step 1 and Step 2:

$y \cdot e^{ax} = \int (e^{mx} \cdot e^{ax}) \, dx + C$

$y \cdot e^{ax} = \int e^{(m+a)x} \, dx + C$

Step 4: Evaluation of the Integral

Applying the standard integral formula $\int e^{kx} \, dx = \frac{e^{kx}}{k} + C$, assuming $m + a \neq 0$:

$y \cdot e^{ax} = \frac{e^{(m+a)x}}{m+a} + C$

Step 5: Algebraic Rearrangement

To isolate $y$, divide the entire equation by $e^{ax}$ (or multiply by $e^{-ax}$):

$y = \frac{e^{(m+a)x}}{e^{ax}(m+a)} + \frac{C}{e^{ax}}$

$y = \frac{e^{mx+ax-ax}}{m+a} + C e^{-ax}$

$y = \frac{e^{mx}}{m+a} + C e^{-ax}$

Note on the case $m = -a$: If $m + a = 0$, then $\frac{dy}{dx} + ay = e^{-ax}$. Multiplying by $e^{ax}$ gives $e^{ax}\frac{dy}{dx} + a e^{ax}y = 1$, which is $\frac{d}{dx}(y e^{ax}) = 1$. Integrating both sides: $y e^{ax} = x + C \implies y = (x+C)e^{-ax}$.

Final Solution: For $m \neq -a$, the general solution is $y = \frac{e^{mx}}{m+a} + Ce^{-ax}$, where $C$ is an arbitrary constant.

Step 1: Setting up the Differential Equation

Let the point $P$ on the curve be $(x, y)$. The equation of the tangent at point $P(x, y)$ is given by:

$Y - y = \frac{dy}{dx}(X - x)$ [Per the point-slope form of a linear equation]

To find the coordinates of points $A$ and $B$, we determine the intercepts of this tangent line:

• For the $x$-intercept (point $A$, where $Y=0$):
  $0 - y = \frac{dy}{dx}(X_A - x) \implies X_A = x - y\frac{dx}{dy}$
  So, $A = (x - y\frac{dx}{dy}, 0)$
• For the $y$-intercept (point $B$, where $X=0$):
  $Y_B - y = \frac{dy}{dx}(0 - x) \implies Y_B = y - x\frac{dy}{dx}$
  So, $B = (0, y - x\frac{dy}{dx})$

Step 2: Applying the Mid-point Condition

Given that $P(x, y)$ is the mid-point of $AB$, we apply the mid-point formula $P = \left(\frac{X_A + X_B}{2}, \frac{Y_A + Y_B}{2}\right)$:

$x = \frac{(x - y\frac{dx}{dy}) + 0}{2} \implies 2x = x - y\frac{dx}{dy} \implies x = -y\frac{dx}{dy}$

$y = \frac{0 + (y - x\frac{dy}{dx})}{2} \implies 2y = y - x\frac{dy}{dx} \implies y = -x\frac{dy}{dx}$

Both conditions yield the same differential equation: $\frac{dy}{dx} = -\frac{y}{x}$.

Step 3: Solving the Differential Equation

We solve by the method of separation of variables [By the Principle of Separable Variables]:

$\int \frac{dy}{y} = -\int \frac{dx}{x}$

$\ln|y| = -\ln|x| + C$

$\ln|y| + \ln|x| = C \implies \ln|xy| = C$

$xy = e^C = k$ (where $k$ is an arbitrary constant)

Step 4: Applying the Boundary Condition

The curve passes through the point $(1, 1)$. Substituting these values into $xy = k$:

$(1)(1) = k \implies k = 1$

Step 5: Final Equation Formulation

The resulting curve is defined by the product of its coordinates being constant.

Final Solution: xy = 1

Step 1: Identification of the Differential Equation Form

The given equation is $\frac{dy}{dx} + \frac{y}{x} = \sin x$. This is a first-order linear ordinary differential equation of the form:

$\frac{dy}{dx} + P(x)y = Q(x)$ [Standard Linear Form]

By comparing the two, we identify:

• $P(x) = \frac{1}{x}$
• $Q(x) = \sin x$

Step 2: Calculating the Integrating Factor (IF)

The Integrating Factor is defined by the formula $IF = e^{\int P(x) dx}$.

$IF = e^{\int \frac{1}{x} dx}$ [Integral of $\frac{1}{x}$ is $\ln|x|$]

$IF = e^{\ln x} = x$ [Property of Logarithms: $e^{\ln a} = a$]

Step 3: Applying the General Solution Formula

The general solution for a linear differential equation is given by:

$y \cdot (IF) = \int (Q(x) \cdot IF) dx + C$

Substituting the values derived:

$y \cdot x = \int (\sin x \cdot x) dx + C$

Step 4: Integration by Parts

To solve $\int x \sin x dx$, we use the Integration by Parts formula: $\int u dv = uv - \int v du$.

Let $u = x \implies du = dx$

Let $dv = \sin x dx \implies v = -\cos x$

$\int x \sin x dx = x(-\cos x) - \int (-\cos x) dx$

$= -x \cos x + \int \cos x dx$

$= -x \cos x + \sin x + C$ [Integral of $\cos x$ is $\sin x$]

Step 5: Final Compilation of the Solution

Equating the result from Step 3 and Step 4:

$xy = -x \cos x + \sin x + C$

To express the solution in terms of $y$, divide the entire equation by $x$:

$y = -\cos x + \frac{\sin x}{x} + \frac{C}{x}$

Visual Representation of the Solution Family

Final Solution: $xy = \sin x - x \cos x + C$ or $y = \frac{\sin x - x \cos x + C}{x}$

Step 1: Identifying the Given Equation

The given equation is an implicit function representing a family of curves:

$Ax^2 + By^2 = 1$ --- (Equation 1)

Here, $A$ and $B$ are arbitrary constants to be eliminated. Since there are two constants, we must differentiate the equation twice to obtain a second-order differential equation [Per the property of eliminating $n$ arbitrary constants in an $n^{th}$ order differential equation].

Step 2: First Order Differentiation

Differentiating Equation 1 with respect to $x$ using the power rule ($\frac{d}{dx}x^n = nx^{n-1}$):

$\frac{d}{dx}(Ax^2) + \frac{d}{dx}(By^2) = \frac{d}{dx}(1)$

$2Ax + 2By\frac{dy}{dx} = 0$

Dividing by 2:

$Ax + By y' = 0$ --- (Equation 2)

Step 3: Second Order Differentiation

Differentiating Equation 2 with respect to $x$ using the product rule $[\frac{d}{dx}(uv) = u'v + uv']$:

$\frac{d}{dx}(Ax) + \frac{d}{dx}(Byy') = \frac{d}{dx}(0)$

$A + B(y \cdot y'' + y' \cdot y') = 0$

$A + B(yy'' + (y')^2) = 0$ --- (Equation 3)

Step 4: Algebraic Elimination of Constants

From Equation 2, we find $A$ in terms of $B$:

$Ax = -Byy' \implies A = -\frac{Byy'}{x}$

Substitute this value of $A$ into Equation 3:

$-\frac{Byy'}{x} + B(yy'' + (y')^2) = 0$

Assuming $B \neq 0$ (otherwise the original equation collapses), divide by $B$:

$-\frac{yy'}{x} + yy'' + (y')^2 = 0$

Step 5: Simplification

Multiply the entire equation by $x$ to clear the denominator:

$-yy' + xyy'' + x(y')^2 = 0$

Rearranging the terms:

$xyy'' + x(y')^2 - yy' = 0$

Final Solution: $xy\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 - y\frac{dy}{dx} = 0$

Step 1: Analyzing the Differential Equation

The given differential equation is: $\tan y \sec^2 x \, dx + \tan x \sec^2 y \, dy = 0$ [This is a first-order, first-degree differential equation.]

Step 2: Separating the Variables

To solve by the method of separation of variables, we isolate terms involving $x$ on one side and terms involving $y$ on the other side. Subtract $\tan x \sec^2 y \, dy$ from both sides: $\tan y \sec^2 x \, dx = -\tan x \sec^2 y \, dy$ [Transposition Property]

Divide both sides by $(\tan x \tan y)$: $\frac{\sec^2 x}{\tan x} \, dx = -\frac{\sec^2 y}{\tan y} \, dy$ [Separation of Variables]

Step 3: Integration

Integrate both sides: $\int \frac{\sec^2 x}{\tan x} \, dx = -\int \frac{\sec^2 y}{\tan y} \, dy$

For the left integral, let $u = \tan x$, then $du = \sec^2 x \, dx$. For the right integral, let $v = \tan y$, then $dv = \sec^2 y \, dy$. This leads to: $\int \frac{1}{u} \, du = -\int \frac{1}{v} \, dv$ [Substitution Rule for Integration]

Integrating gives: $\ln|\tan x| = -\ln|\tan y| + C$ [Integration of $\frac{1}{f(x)}f'(x) \, dx = \ln|f(x)| + C$]

Step 4: Simplifying the Solution

Rearrange the logarithmic terms: $\ln|\tan x| + \ln|\tan y| = C$ Using the logarithmic property $\ln(a) + \ln(b) = \ln(ab)$: $\ln|\tan x \cdot \tan y| = C$

Exponentiate both sides to eliminate the natural logarithm: $|\tan x \cdot \tan y| = e^C$ Since $e^C$ is an arbitrary constant, let $e^C = k$ (where $k > 0$): $\tan x \tan y = k$

Visual Representation (Conceptual Map of Slope Field)

Conclusion

The general solution describes a family of hyperbolic-like curves where the product of the tangents of the variables is constant.

Final Solution: $\tan x \tan y = C$

Step 1: Identifying the Differential Equation

The given differential equation is:

$[1 + (\frac{dy}{dx})^2]^{\frac{3}{2}} = \frac{d^2y}{dx^2}$

Step 2: Eliminating Fractional Exponents

To determine the degree of a differential equation, all derivatives must be expressed as powers of non-negative integers [Per the definition of the degree of a differential equation, which requires the equation to be a polynomial in terms of its derivatives].

To remove the exponent of $\frac{3}{2}$, we square both sides of the equation:

$([1 + (\frac{dy}{dx})^2]^{\frac{3}{2}})^2 = (\frac{d^2y}{dx^2})^2$

$[1 + (\frac{dy}{dx})^2]^3 = (\frac{d^2y}{dx^2})^2$

Step 3: Determination of Order

The order of a differential equation is the order of the highest derivative present in the equation [By Definition of Order of Differential Equation].

• The derivatives present are $\frac{dy}{dx}$ (first order) and $\frac{d^2y}{dx^2}$ (second order).
• The highest derivative is $\frac{d^2y}{dx^2}$.
• Therefore, the Order = 2.

Step 4: Determination of Degree

The degree of a differential equation is the power of the highest order derivative when the equation is expressed as a polynomial in its derivatives [By Definition of Degree of Differential Equation].

• In the simplified equation $[1 + (\frac{dy}{dx})^2]^3 = (\frac{d^2y}{dx^2})^2$, the highest order derivative is $\frac{d^2y}{dx^2}$.
• The exponent of the highest order derivative $\frac{d^2y}{dx^2}$ is $2$.
• Therefore, the Degree = 2.

Step 5: Visual Representation of Derivative Hierarchy

Final Solution: The order of the differential equation is 2 and the degree is 2.

Step 1: Identifying the General Solution and Parameters

The given function is a general solution of a linear homogeneous second-order differential equation with constant coefficients, represented by:

$y = ae^{mx} + be^{-mx}$

Here, $a$ and $b$ are arbitrary constants. To derive the corresponding differential equation, we must eliminate these two constants by successive differentiation. [By the Fundamental Theorem of Ordinary Differential Equations, a solution with $n$ arbitrary constants corresponds to an $n^{th}$ order differential equation].

Step 2: First-Order Differentiation

Differentiate $y$ with respect to $x$ using the chain rule $\frac{d}{dx}[e^{kx}] = ke^{kx}$:

$y' = \frac{dy}{dx} = \frac{d}{dx}(ae^{mx}) + \frac{d}{dx}(be^{-mx})$

$y' = ame^{mx} - bme^{-mx}$ --- (Equation 1)

Step 3: Second-Order Differentiation

Differentiate $y'$ with respect to $x$ once more:

$y'' = \frac{d^2y}{dx^2} = \frac{d}{dx}(ame^{mx}) - \frac{d}{dx}(bme^{-mx})$

$y'' = am^2e^{mx} + bm^2e^{-mx}$ --- (Equation 2)

Step 4: Algebraic Elimination of Constants

Observe Equation 2 and factor out the common constant $m^2$:

$y'' = m^2(ae^{mx} + be^{-mx})$

Since the original expression provided is $y = ae^{mx} + be^{-mx}$, we substitute this directly into the equation above:

$y'' = m^2 y$

Step 5: Final Formulation of the Differential Equation

Rearranging the equation into the standard form $f(y'', y', y) = 0$:

$y'' - m^2 y = 0$

Or, expressed in Leibniz notation:

$\frac{d^2y}{dx^2} - m^2y = 0$

Final Solution: The differential equation satisfied by the given function is $\frac{d^2y}{dx^2} - m^2y = 0$.

Step 1: Identifying the Differential Equation

The given differential equation is:

$y \frac{dy}{dx} + x = c$

where $c$ is an arbitrary constant. This is a first-order, first-degree differential equation. [By definition of Separable Variables differential equations].

Step 2: Separation of Variables

Rearrange the equation to isolate variables $x$ and $y$ on opposite sides of the equality sign:

$y \, dy = (c - x) \, dx$

Step 3: Integration

Integrate both sides with respect to their respective variables:

$\int y \, dy = \int (c - x) \, dx$

Applying the Power Rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$:

$\frac{y^2}{2} = cx - \frac{x^2}{2} + K$

where $K$ is the constant of integration. [Per the Fundamental Theorem of Calculus].

Step 4: Algebraic Manipulation to Standard Conic Form

Multiply the entire equation by 2 to clear the denominators:

$y^2 = 2cx - x^2 + 2K$

$x^2 - 2cx + y^2 = 2K$

Complete the square for the $x$ terms:

$(x^2 - 2cx + c^2) + y^2 = 2K + c^2$

$(x - c)^2 + y^2 = (constant)^2$

Let $R^2 = 2K + c^2$. The equation simplifies to:

$(x - c)^2 + (y - 0)^2 = R^2$

Step 5: Geometric Interpretation

The derived equation is in the standard form of a circle $(x - h)^2 + (y - k)^2 = r^2$, where the center is $(h, k)$ and the radius is $r$.

Here, the center of the circle is $(c, 0)$, which lies on the x-axis, and the radius is $R = \sqrt{2K + c^2}$.

Visual Representation

Final Solution: The differential equation $y\frac{dy}{dx} + x = c$ represents a family of circles with centers on the x-axis at $(c, 0)$.

Step 1: Standardizing the Differential Equation

The given equation is $dy = \cos x (2 - y \csc x) dx$. To solve this, we express it in the standard linear form of a first-order differential equation: $\frac{dy}{dx} + P(x)y = Q(x)$.

Expand the right-hand side:

$dy = (2 \cos x - y \cos x \csc x) dx$

Since $\csc x = \frac{1}{\sin x}$, then $\cos x \csc x = \frac{\cos x}{\sin x} = \cot x$.

$dy = (2 \cos x - y \cot x) dx$

Dividing by $dx$ and rearranging terms to isolate $y$:

$\frac{dy}{dx} + y \cot x = 2 \cos x$ [Standard Linear Form: $\frac{dy}{dx} + P(x)y = Q(x)$]

Step 2: Calculating the Integrating Factor (IF)

The Integrating Factor is defined as $IF = e^{\int P(x) dx}$.

Here, $P(x) = \cot x$.

$IF = e^{\int \cot x dx}$

$IF = e^{\ln |\sin x|}$ [Integral of $\cot x$ is $\ln |\sin x|$]

$IF = \sin x$

Step 3: General Solution Formulation

The general solution is given by $y \cdot (IF) = \int (Q(x) \cdot IF) dx + C$.

$y \sin x = \int (2 \cos x \cdot \sin x) dx + C$

Apply the trigonometric identity $2 \sin x \cos x = \sin(2x)$:

$y \sin x = \int \sin(2x) dx + C$

$y \sin x = -\frac{\cos(2x)}{2} + C$ [Integral of $\sin(ax)$ is $-\frac{\cos(ax)}{a}$]

Step 4: Applying Initial Conditions

We are given $y = 2$ at $x = \frac{\pi}{2}$.

$2 \sin(\frac{\pi}{2}) = -\frac{\cos(2 \cdot \frac{\pi}{2})}{2} + C$

$2(1) = -\frac{\cos(\pi)}{2} + C$

$2 = -\frac{(-1)}{2} + C$

$2 = \frac{1}{2} + C$

$C = 2 - 0.5 = \frac{3}{2}$

Step 5: Final Equation Consolidation

Substitute $C = \frac{3}{2}$ into the general solution:

$y \sin x = -\frac{\cos(2x)}{2} + \frac{3}{2}$

Using the identity $\cos(2x) = 1 - 2\sin^2 x$:

$y \sin x = -\frac{1 - 2\sin^2 x}{2} + \frac{3}{2}$

$y \sin x = -\frac{1}{2} + \sin^2 x + \frac{3}{2}$

$y \sin x = \sin^2 x + 1$

$y = \sin x + \csc x$

Final Solution: $y = \sin x + \csc x$

Step 1: Standardizing the Differential Equation

The given equation is $x \frac{dy}{dx} = y (\log y - \log x + 1)$. Rearranging the terms into a standard form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$ [By Algebraic Manipulation]:

$\frac{dy}{dx} = \frac{y}{x} \left( \log\left(\frac{y}{x}\right) + 1 \right)$

Step 2: Substitution for Homogeneous Form

Since the right-hand side is a function of the ratio $\frac{y}{x}$, we introduce the substitution $v = \frac{y}{x}$, which implies $y = vx$. Differentiating with respect to $x$ [By Product Rule]:

$\frac{dy}{dx} = v + x \frac{dv}{dx}$

Substituting these into the original equation:

$v + x \frac{dv}{dx} = v (\log v + 1)$

$v + x \frac{dv}{dx} = v \log v + v$

Step 3: Separating Variables

Subtracting $v$ from both sides simplifies the expression to a separable differential equation [By Principle of Separation of Variables]:

$x \frac{dv}{dx} = v \log v$

$\frac{dv}{v \log v} = \frac{dx}{x}$

Step 4: Integration

Integrate both sides: $\int \frac{1}{v \log v} dv = \int \frac{1}{x} dx$

For the left integral, let $u = \log v$, then $du = \frac{1}{v} dv$ [By Integration by Substitution]:

$\int \frac{1}{u} du = \int \frac{1}{x} dx$

$\log |u| = \log |x| + C$

Substituting $u = \log v$ back into the equation:

$\log |\log v| = \log |x| + C$

Step 5: Back-Substitution and Simplification

Replace $v$ with $\frac{y}{x}$ and use the property of logarithms $\log |A| = \log |B| + C \implies A = K \cdot B$ where $K = e^C$:

$\log \left| \log \left( \frac{y}{x} \right) \right| = \log |x| + \log |K|$

$\log \left| \log \left( \frac{y}{x} \right) \right| = \log |Kx|$

$\log \left( \frac{y}{x} \right) = Cx$

$\frac{y}{x} = e^{Cx}$

$y = x e^{Cx}$

Visual Representation of the Solution Path

Final Solution: $y = x e^{Cx}$ where $C$ is an arbitrary constant.

Step 1: Standardizing the Differential Equation

The given linear differential equation is of the form:

$\cos x \frac{dy}{dx} + y \sin x = 1$

To identify the integrating factor, we must transform the equation into the standard linear form: $\frac{dy}{dx} + P(x)y = Q(x)$. Dividing the entire equation by $\cos x$ (assuming $\cos x \neq 0$):

$\frac{dy}{dx} + \left( \frac{\sin x}{\cos x} \right) y = \frac{1}{\cos x}$

[By algebraic division property]

Step 2: Identifying P(x) and Q(x)

Comparing the simplified equation $\frac{dy}{dx} + y \tan x = \sec x$ with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we obtain:

• $P(x) = \tan x$
• $Q(x) = \sec x$

Step 3: Calculating the Integrating Factor (IF)

The formula for the integrating factor of a linear differential equation is defined as $IF = e^{\int P(x) dx}$.

Substituting $P(x) = \tan x$:

$IF = e^{\int \tan x \, dx}$

[By the definition of Integrating Factor for linear ODEs]

Step 4: Integration and Simplification

Using the standard integral identity $\int \tan x \, dx = \ln|\sec x|$:

$IF = e^{\ln|\sec x|}$

Applying the logarithmic identity $e^{\ln(f(x))} = f(x)$:

$IF = \sec x$

[By the property of logarithms and exponentials]

Visual Representation of the Integration Step

Final Solution: The integrating factor of the differential equation is $\sec x$.

Step 1: Identifying the Differential Equation Type

The given equation is $\frac{dy}{dx} + 2xy = y^2$. This is a Bernoulli differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$, where $n = 2$. To linearize this equation, we divide both sides by $y^2$:

$y^{-2} \frac{dy}{dx} + 2xy^{-1} = 1$ [Dividing by $y^n$ to isolate the non-linear term]

Step 2: Substitution for Linearization

Let $v = y^{1-n} = y^{1-2} = y^{-1}$.

Differentiating both sides with respect to $x$ using the chain rule:

$\frac{dv}{dx} = -1 \cdot y^{-2} \frac{dy}{dx} \implies - \frac{dv}{dx} = y^{-2} \frac{dy}{dx}$

Substituting this into our equation from Step 1:

$-\frac{dv}{dx} + 2xv = 1$

Multiplying by $-1$ to standardize the linear form:

$\frac{dv}{dx} - 2xv = -1$ [Standard linear form: $\frac{dv}{dx} + P(x)v = Q(x)$]

Step 3: Finding the Integrating Factor (IF)

Here, $P(x) = -2x$ and $Q(x) = -1$.

$IF = e^{\int P(x) dx} = e^{\int -2x dx} = e^{-x^2}$ [Definition of Integrating Factor for linear ODEs]

Step 4: Integrating the Linear Equation

The solution is given by $v \cdot (IF) = \int Q(x) \cdot (IF) dx + C$.

$v \cdot e^{-x^2} = \int -1 \cdot e^{-x^2} dx + C$

Note: $\int e^{-x^2} dx$ is not expressible in elementary functions; however, we retain the form using the error function $erf(x)$ or maintain the integral notation if required by standard curriculum conventions.

$v e^{-x^2} = -\int e^{-x^2} dx + C$

Substituting back $v = y^{-1} = \frac{1}{y}$:

$\frac{1}{y} e^{-x^2} = -\int e^{-x^2} dx + C$

Step 5: Visual Representation of the Vector Field

Step 6: Final Algebraic Arrangement

To isolate $y$, we invert the expression:

$y = \frac{e^{-x^2}}{C - \int e^{-x^2} dx}$

Final Solution: $y = \frac{e^{-x^2}}{C - \int_{0}^{x} e^{-t^2} dt}$, where $C$ is an arbitrary constant.

Analytical Assessment of the Statement

The statement posits that for a differential equation of the form $\frac{dx}{dy} = g(x, y)$, where $g(x, y)$ is a homogeneous function of degree zero, the appropriate substitution is $x = vy$.

Step 1: Theoretical Definition of Homogeneous Differential Equations

A differential equation of the form $\frac{dx}{dy} = g(x, y)$ is defined as homogeneous of degree zero if the function $g(x, y)$ satisfies the condition:

$g(\lambda x, \lambda y) = \lambda^0 g(x, y) = g(x, y)$ for any non-zero constant $\lambda$.

[Per the definition of homogeneous functions of degree $n$, where $n=0$].

Step 2: Justification of the Substitution Method

When a differential equation is expressed as $\frac{dx}{dy} = g(x, y)$, the independent variable is $y$ and the dependent variable is $x$. To transform this into a separable variable form, we introduce a new dependent variable $v$ such that:

$x = vy$

Differentiating both sides with respect to $y$ using the Product Rule [Ref: $\frac{d}{dy}(uv) = u\frac{dv}{dy} + v\frac{du}{dy}$]:

$\frac{dx}{dy} = v(1) + y\frac{dv}{dy}$

$\frac{dx}{dy} = v + y\frac{dv}{dy}$

Step 3: Verification of Transformation

Since $g(x, y)$ is homogeneous of degree zero, we can express it as a function of the ratio $\frac{x}{y}$:

$g(x, y) = f\left(\frac{x}{y}\right)$

Substituting $x = vy$ into the original equation:

$v + y\frac{dv}{dy} = f(v)$

$y\frac{dv}{dy} = f(v) - v$

$\frac{dv}{f(v) - v} = \frac{dy}{y}$

This result confirms that the substitution $x = vy$ effectively separates the variables $v$ and $y$, allowing for integration.

Conclusion

The substitution $x = vy$ is the standard and correct mathematical procedure for solving homogeneous differential equations of the form $\frac{dx}{dy} = g(x, y)$.

Final Solution: True

Step 1: Analyzing the Differential Equation

Given the first-order differential equation:

$e^x \cos y \, dx - e^x \sin y \, dy = 0$

This is a first-order, first-degree differential equation. To solve it, we must separate the variables $x$ and $y$ [Per the Variable Separable Method].

Step 2: Separation of Variables

Rearrange the terms to isolate the differentials $dx$ and $dy$:

$e^x \cos y \, dx = e^x \sin y \, dy$

Assuming $e^x \neq 0$ (which is always true for real $x$), we divide both sides by $e^x$:

$\cos y \, dx = \sin y \, dy$

Now, separate the variables such that $x$ terms are with $dx$ and $y$ terms are with $dy$:

$dx = \frac{\sin y}{\cos y} \, dy$

$dx = \tan y \, dy$

Step 3: Integration of Both Sides

Integrate both sides with respect to their respective variables:

$\int 1 \, dx = \int \tan y \, dy$

Using the standard integral formula $\int \tan y \, dy = \ln|\sec y| + C$ or $-\ln|\cos y| + C$ [Per Integral Calculus Theorems]:

$x = \ln|\sec y| + C$

Step 4: Algebraic Simplification

We can represent the constant $C$ as $\ln|k|$ to simplify the logarithmic expression:

$x = \ln|\sec y| + \ln|k|$

$x = \ln|k \sec y|$

Applying the exponential function to both sides:

$e^x = k \sec y$

Since $\sec y = \frac{1}{\cos y}$, we can write:

$e^x = \frac{k}{\cos y}$

$e^x \cos y = k$

Visual Representation of the Solution Family

Step 5: Final Verification

Differentiating $e^x \cos y = C$ with respect to $x$:

$\frac{d}{dx}(e^x \cos y) = \frac{d}{dx}(C)$

$e^x \cos y + e^x (-\sin y) \frac{dy}{dx} = 0$

$e^x \cos y - e^x \sin y \frac{dy}{dx} = 0$

Multiplying by $dx$: $e^x \cos y \, dx - e^x \sin y \, dy = 0$. This confirms the derivation is accurate.

Final Solution: $e^x \cos y = C$

Step 1: Identifying the Differential Equation Components

The given differential equation is:

$\left(\frac{d^2y}{dx^2}\right)^2 + \left(\frac{dy}{dx}\right)^2 = x \sin\left(\frac{dy}{dx}\right)$

To determine the degree of a differential equation, we must analyze the structure of the equation regarding its highest-order derivative.

Step 2: Defining Order and Degree

• Order: The order of a differential equation is the order of the highest derivative present in the equation [Per the standard definition of order in Ordinary Differential Equations]. Here, the highest derivative is $\frac{d^2y}{dx^2}$, which is of order 2.
• Degree: The degree of a differential equation is defined as the power of the highest-order derivative, provided the equation is a polynomial equation in terms of its derivatives [Per the standard definition of degree].

Step 3: Analyzing the Transcendental Term

A differential equation must be a polynomial in its derivatives (i.e., $\frac{dy}{dx}, \frac{d^2y}{dx^2}, \dots, \frac{d^ny}{dx^n}$) for the degree to be defined. If a derivative appears as an argument of a transcendental function—such as $\sin(\cdot)$, $\cos(\cdot)$, $e^{(\cdot)}$, or $\log(\cdot)$—the equation is non-polynomial [By the Theory of Transcendental Functions in Differential Equations].

In the expression $x \sin\left(\frac{dy}{dx}\right)$, the term $\frac{dy}{dx}$ is inside the sine function. Using the Taylor series expansion for the sine function:

$\sin\left(\frac{dy}{dx}\right) = \frac{dy}{dx} - \frac{(\frac{dy}{dx})^3}{3!} + \frac{(\frac{dy}{dx})^5}{5!} - \dots$

Substituting this back into the equation reveals an infinite series of powers of the derivative $\frac{dy}{dx}$.

Step 4: Evaluating the Mathematical Conclusion

Criterion	Analysis
Highest Order Derivative	$\frac{d^2y}{dx^2}$ (Order 2)
Polynomial Nature	Non-polynomial due to $\sin(\frac{dy}{dx})$
Degree Status	Undefined

Because the equation involves the derivative $\frac{dy}{dx}$ in a transcendental function, it cannot be expressed as a polynomial in terms of its derivatives. Consequently, the degree of this differential equation does not exist.

Final Solution: The degree of the differential equation is not defined.

Step 1: Rearranging the Differential Equation

The given differential equation is:

$\frac{dy}{dx} = e^{x-y} + x^2 e^{-y}$

Utilizing the laws of exponents, specifically $a^{m-n} = \frac{a^m}{a^n}$, we rewrite the expression as:

$\frac{dy}{dx} = \frac{e^x}{e^y} + \frac{x^2}{e^y}$ [Property of Exponents: $e^{x-y} = e^x \cdot e^{-y} = \frac{e^x}{e^y}$]

Factoring out $\frac{1}{e^y}$ from the right-hand side:

$\frac{dy}{dx} = e^{-y}(e^x + x^2)$

Step 2: Separation of Variables

To solve the equation, we move all terms involving $y$ to the left side and all terms involving $x$ to the right side:

$e^y \, dy = (e^x + x^2) \, dx$ [Method of Variable Separation]

Step 3: Integration of Both Sides

We integrate both sides with respect to their respective variables:

$\int e^y \, dy = \int (e^x + x^2) \, dx$

Applying the standard integration formulas $\int e^u du = e^u + C$ and $\int x^n dx = \frac{x^{n+1}}{n+1} + C$:

$e^y = \int e^x \, dx + \int x^2 \, dx$

$e^y = e^x + \frac{x^3}{3} + C$ [Where $C$ is the constant of integration]

Step 4: Isolating $y$ (Optional/Alternative representation)

To isolate $y$, we take the natural logarithm ($\ln$) of both sides:

$y = \ln\left(e^x + \frac{x^3}{3} + C\right)$

Visual Representation of the Solution Family

Final Solution: The general solution to the differential equation is $e^y = e^x + \frac{x^3}{3} + C$ or $y = \ln\left(e^x + \frac{x^3}{3} + C\right)$.

Step 1: Analyzing the Differential Equation

The given differential equation is: $y \, dx + (x + xy) \, dy = 0$.

To determine the solution, we rearrange the equation into a standard form. We separate the variables or attempt to express it as an exact equation. Rearranging the terms:

$y \, dx = -(x + xy) \, dy$

$\frac{dx}{dy} = -\frac{x(1 + y)}{y}$ [By algebraic rearrangement of terms]

Step 2: Separation of Variables

We group all terms containing $x$ on one side and $y$ on the other:

$\frac{1}{x} \, dx = -\frac{1 + y}{y} \, dy$

$\frac{1}{x} \, dx = -\left( \frac{1}{y} + 1 \right) \, dy$

Step 3: Integration

Integrate both sides with respect to their respective variables:

$\int \frac{1}{x} \, dx = -\int \left( \frac{1}{y} + 1 \right) \, dy$

Applying standard integration formulas [$\int \frac{1}{u} du = \ln|u| + C$]:

$\ln|x| = -(\ln|y| + y) + C_1$

$\ln|x| = -\ln|y| - y + C_1$

Step 4: Simplifying the Expression

Move the logarithmic terms to one side:

$\ln|x| + \ln|y| = -y + C_1$

Using the logarithmic identity $\ln A + \ln B = \ln(AB)$:

$\ln|xy| = -y + C_1$

To eliminate the natural logarithm, exponentiate both sides:

$|xy| = e^{-y + C_1}$

$xy = \pm e^{C_1} \cdot e^{-y}$

Let $C = \pm e^{C_1}$, which is an arbitrary constant:

$xy = Ce^{-y}$

Visualization of the General Solution

Comparison of Formats

Format	Result
Logarithmic form	$\ln|xy| + y = C$
Exponential form	$xy = Ce^{-y}$

Final Solution: $xy = Ce^{-y}$ or $\ln|xy| + y = C$

Step 1: Analyzing the Differential Equation

The given differential equation is:

$\frac{dy}{dx} = \frac{x + 2y}{x}$

This is a first-order linear homogeneous differential equation of the form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$. We can rewrite the equation as:

$\frac{dy}{dx} = 1 + 2\left(\frac{y}{x}\right)$

Step 2: Solving via Substitution

Let $y = vx$. Then, by the product rule for differentiation, $\frac{dy}{dx} = v + x\frac{dv}{dx}$ [Per Product Rule]. Substituting this into the original equation:

$v + x\frac{dv}{dx} = 1 + 2v$

$x\frac{dv}{dx} = 1 + v$

Step 3: Separation of Variables and Integration

Rearranging the terms to separate variables:

$\int \frac{1}{1 + v} dv = \int \frac{1}{x} dx$

Integrating both sides [Per Fundamental Theorem of Calculus]:

$\ln|1 + v| = \ln|x| + C$

Let $C = \ln|A|$ for mathematical convenience:

$\ln|1 + v| = \ln|Ax|$

$1 + v = Ax$

Step 4: Back-substitution and Verification

Substitute $v = \frac{y}{x}$ back into the equation:

$1 + \frac{y}{x} = Ax$

Multiply the entire equation by $x$:

$x + y = Ax^2$

Step 5: Comparative Analysis

The solution derived from the analytical process is $x + y = Ax^2$ (where $A$ is an arbitrary constant). The statement provided in the question claims the solution is $x + y = kx^2$.

Comparing the two results:

Component	Analytical Result	Given Claim
General Solution	$x + y = Ax^2$	$x + y = kx^2$

Both expressions represent the same family of curves, as $A$ and $k$ are arbitrary constants representing the constant of integration.

Final Solution: True. The given expression $x + y = kx^2$ is indeed a valid representation of the general solution for the differential equation $\frac{dy}{dx} = \frac{x + 2y}{x}$.

Step 1: Standardizing the Linear Differential Equation

The given differential equation is: $x \frac{dy}{dx} - y = x^4 - 3x$ A first-order linear differential equation is standardly represented as: $\frac{dy}{dx} + P(x)y = Q(x)$ [Standard form of Linear Differential Equation] To transform the given equation into the standard form, divide every term by the coefficient of $\frac{dy}{dx}$, which is $x$ ($x \neq 0$): $\frac{dy}{dx} - \frac{1}{x}y = \frac{x^4 - 3x}{x}$ Simplifying the right-hand side: $\frac{dy}{dx} - \frac{1}{x}y = x^3 - 3$

Step 2: Identifying P(x) and Q(x)

By comparing $\frac{dy}{dx} - \frac{1}{x}y = x^3 - 3$ with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we identify: $P(x) = -\frac{1}{x}$ $Q(x) = x^3 - 3$

Step 3: Calculating the Integrating Factor (I.F.)

The formula for the Integrating Factor (I.F.) is given by: $I.F. = e^{\int P(x) \, dx}$ [Definition of Integrating Factor] Substituting the value of $P(x)$: $I.F. = e^{\int (-\frac{1}{x}) \, dx}$ Perform the integration: $\int -\frac{1}{x} \, dx = -\ln|x|$ [Per the Integral formula $\int \frac{1}{x} dx = \ln|x|$] Thus: $I.F. = e^{-\ln|x|}$

Step 4: Algebraic Simplification

Using the logarithmic property $n \ln(a) = \ln(a^n)$: $-\ln|x| = \ln(|x|^{-1}) = \ln(\frac{1}{|x|})$ Substituting back into the exponent: $I.F. = e^{\ln(\frac{1}{|x|})}$ Applying the identity $e^{\ln(f(x))} = f(x)$: $I.F. = \frac{1}{x}$

Step 5: Graphical Representation of the Function x⁻¹

Final Solution: The Integrating Factor is $\frac{1}{x}$.

Step 1: Identification of the Differential Equation Type

The given equation is of the standard linear first-order differential equation form:

$\frac{dy}{dx} + P(x)y = Q(x)$

[By the classification of First-Order Linear Differential Equations, where $P(x)$ and $Q(x)$ are functions of $x$ alone.]

Step 2: Extraction of the Coefficient Function $P(x)$

Comparing the provided equation $\frac{dy}{dx} + (1 + \frac{1}{x})y = Q(x)$ with the standard form, we identify the term $P(x)$:

$P(x) = 1 + \frac{1}{x}$

Step 3: Derivation of the Integrating Factor (IF) Formula

The Integrating Factor ($IF$) for a linear differential equation of the form $\frac{dy}{dx} + Py = Q$ is defined by the exponential integral formula:

$IF = e^{\int P(x) \, dx}$

[Per the definition of the integrating factor in the context of linear differential equations.]

Step 4: Evaluating the Integral of $P(x)$

Substitute $P(x)$ into the exponent:

$\int P(x) \, dx = \int (1 + \frac{1}{x}) \, dx$

Applying the linearity property of integration:

$\int 1 \, dx + \int \frac{1}{x} \, dx = x + \ln|x|$

[Using the standard integration formulas $\int x^n dx = \frac{x^{n+1}}{n+1}$ and $\int \frac{1}{x} dx = \ln|x| + C$.]

Step 5: Final Calculation of the Integrating Factor

Substitute the evaluated integral back into the expression for $IF$:

$IF = e^{x + \ln|x|}$

Utilizing the laws of exponents ($e^{a+b} = e^a \cdot e^b$):

$IF = e^x \cdot e^{\ln|x|}$

Applying the identity $e^{\ln(f(x))} = f(x)$:

$IF = e^x \cdot x$

$IF = x e^x$

Visual Representation of the Integrating Factor Growth

Final Solution: The integrating factor of the differential equation is $x e^x$.