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Q77(viii):
State True or False: Differential equation representing the family of curves $y = e^x (A\cos x + B\sin x)$ is $\frac{d^2y}{dx^2} – 2\frac{dy}{dx} + 2y = 0$.

Solution :

Step 1: Analyzing the General Solution

The given family of curves is defined by the equation:

$y = e^x(A\cos x + B\sin x)$ [Equation 1]

Where $A$ and $B$ are arbitrary constants. Since there are two independent arbitrary constants, the corresponding differential equation must be of the second order [Per the theory of Differential Equations: The order of a differential equation represents the number of arbitrary constants in its primitive].

Step 2: First Order Differentiation

Differentiating $y$ with respect to $x$ using the Product Rule, $\frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx}$:

$\frac{dy}{dx} = \frac{d}{dx}[e^x] \cdot (A\cos x + B\sin x) + e^x \cdot \frac{d}{dx}[A\cos x + B\sin x]$

$\frac{dy}{dx} = e^x(A\cos x + B\sin x) + e^x(-A\sin x + B\cos x)$ [Equation 2]

Notice that the first term is equivalent to the original $y$ from Equation 1. Thus, we can simplify:

$\frac{dy}{dx} = y + e^x(B\cos x - A\sin x)$ [Equation 3]

Step 3: Second Order Differentiation

Differentiating Equation 3 with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{dy}{dx} + \frac{d}{dx}[e^x(B\cos x - A\sin x)]$

Applying the Product Rule again to the second term:

$\frac{d^2y}{dx^2} = \frac{dy}{dx} + [e^x(B\cos x - A\sin x) + e^x(-B\sin x - A\cos x)]$

Group the terms to extract original variables:

$\frac{d^2y}{dx^2} = \frac{dy}{dx} + [e^x(B\cos x - A\sin x)] - e^x(A\cos x + B\sin x)$

Step 4: Algebraic Substitution

From Equation 3, we know $e^x(B\cos x - A\sin x) = \frac{dy}{dx} - y$.

From Equation 1, we know $e^x(A\cos x + B\sin x) = y$.

Substitute these into the second derivative expression:

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

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

Step 5: Final Arrangement

Rearrange the terms to equate to zero:

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

Visual Verification

Family of Curves: y = e^x(A cos x + B sin x) Differential Operator: D^2 - 2D + 2 = 0

Final Solution: True. The given differential equation correctly represents the family of curves.


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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