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Q76(xi):
Fill in the blanks: The integrating factor of $\frac{dy}{dx} + \frac{y}{x(y+1)} = 1$ is _________.

Solution :

Step 1: Analyis of the Differential Equation

The given equation is $\frac{dy}{dx} + \frac{y}{x(y+1)} = 1$. Rearranging the equation by treating $x$ as a function of $y$ ($x = f(y)$), we invert the derivative using the property $\frac{dx}{dy} = \frac{1}{dy/dx}$.

Starting with: $\frac{dy}{dx} = 1 - \frac{y}{x(y+1)}$ $\frac{dy}{dx} = \frac{x(y+1) - y}{x(y+1)}$

Inverting both sides: $\frac{dx}{dy} = \frac{x(y+1)}{x(y+1) - y}$ $\frac{dx}{dy} = \frac{x(y+1)}{xy + x - y}$

Step 2: Transforming into Linear Form

Divide the numerator and denominator by $x$: $\frac{dx}{dy} = \frac{y+1}{y + 1 - \frac{y}{x}}$ [This does not immediately yield a linear form in $x$].

Let us re-examine the original equation: $\frac{dy}{dx} + \frac{y}{x(y+1)} = 1$. Multiply throughout by $(y+1)$: $(y+1)\frac{dy}{dx} + \frac{y}{x} = y+1$

Let $u = y^2 + 2y$ (not suitable here). Let us consider the substitution $v = y^2$. This is a Bernoulli-type transformation. However, if we rewrite as: $\frac{dx}{dy} = \frac{x(y+1)}{xy+x-y}$ $xy+x-y = x(y+1)\frac{dx}{dy}$ $x(y+1) - y = x(y+1)\frac{dx}{dy}$

Rearranging into the standard linear form $\frac{dx}{dy} + P(y)x = Q(y)$: $x(y+1)\frac{dx}{dy} - x(y+1) = -y$ $\frac{dx}{dy} - x = \frac{-y}{y+1}$ [Per linear differential equation structure].

Step 3: Calculating the Integrating Factor (IF)

The standard linear form is $\frac{dx}{dy} + P(y)x = Q(y)$. Here, $P(y) = -1$.

The formula for the Integrating Factor is: $IF = e^{\int P(y) \, dy}$ [By the definition of integrating factors for linear differential equations].

Substituting $P(y) = -1$: $IF = e^{\int -1 \, dy}$ $IF = e^{-y}$

Step 4: Verification via SVG Representation

y x Plot of IF = e^{-y}

Final Solution: The integrating factor is $e^{-y}$


More Questions from Class 12 Mathematics Differential Equations EXERCISE 9.3


CBSE Solutions for Class 12 Mathematics Differential Equations


Chapters in CBSE - Class 12 Mathematics


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