y=e^{x}(A x+B x) is the solution of the differential equation

Correct Answer is : d2ydx2−2dydx+2y=0{d² y}{d x²}-2 d y/d x+2 y=0dx2d2y−2dxdy+2y=0

x² {d² y}{d x²}+≤ft(1+y²)=0

x {d² y}{d x²}-2 d y/d x+2 y=0

Correct Answer is : d2ydx2−2dydx+2y=0{d² y}{d x²}-2 d y/d x+2 y=0dx2d2y−2dxdy+2y=0

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MHTCET 25 Apr 2025 Shift 1 Paper

Question : 138 of 150

Marks: +1, -0

y=e^{x}(A \cos x+B \sin x)

is the solution of the differential equation

Solution:

Concept:

Given a function

y

, differentiate it twice and eliminate

A

and

B

to obtain the corresponding differential equation.

Explanation:

Let

y = e^{x}(A \cos x + B \sin x)

Differentiating with respect to

x

\frac{dy}{dx} = e^{x}(A \cos x + B \sin x) + e^{x}(-A \sin x + B \cos x)

\frac{dy}{dx} = y + e^{x}(B \cos x - A \sin x) \quad \dots (1)

Differentiate (1) again:

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

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

Using (1),

e^{x}(B \cos x - A \sin x) = \frac{dy}{dx} - y

, and

e^{x}(A \cos x + B \sin x) = y

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

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

Rearranging:

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

Answer:

Option C:

\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0

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