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  • If y = e-x (A cos x + B sin x), then y is a solution of A.d^2y/dx^2+2dy/dx=0

if \mathrm{y}=\mathrm{e}^{-\mathrm{x}}(\mathrm{A} \cos \mathrm{x}+\mathrm{B} \sin \mathrm{x}) then y is a solution of 

\\A. \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+2 \frac{\mathrm{dy}}{\mathrm{dx}}=0 \\\\\mathrm{B}. \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0 \\\\C. \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+2 y=0 \\\\D. \frac{d^{2} y}{d x^{2}}+2 y=0

Answers (1)

Solution

If \mathrm{y}=\mathrm{f}(\mathrm{x}) is a solution of differential equation, then differentiating it will give the same differential equation.

Differentiate the differential equation twice. Twice because all the options have order as 2 and also because there are two constants A and B
y=e^{-x}(A \cos x+B \sin x)
Differentiating using product rule
\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{e}^{-\mathrm{x}}(\mathrm{Acosx}+\mathrm{B} \sin \mathrm{x})+\mathrm{e}^{-\mathrm{x}}(-\mathrm{A} \sin \mathrm{x}+\mathrm{B} \cos \mathrm{x})
But e^{-x}(A \cos x+B \sin x)=y

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{y}+\mathrm{e}^{-\mathrm{x}}(-\mathrm{Asinx}+\mathrm{Bcosx})
Differentiating again with respect to x,
\\ \Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{d y}{d x}-e^{-x}(-A \sin x+B \cos x)+e^{-x}(-A \cos x-B \sin x) \\ \Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{d y}{d x}-e^{-x}(-A \sin x+B \cos x)-e^{-x}(A \cos x+B \sin x)
But e^{-x}(A \cos x+B \sin x)=y
\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{e}^{-\mathrm{x}}(-\mathrm{Asinx}+\mathrm{B} \cos \mathrm{x})-\mathrm{y}
Also,
\frac{d y}{d x}=-y+e^{-x}(-A \sin x+B \cos x)
Means,
\\ e^{-x}(-A \sin x+B \cos x)=\frac{d y}{d x}+y \\ \quad \Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{d y}{d x}-\left(\frac{d y}{d x}+y\right)-y

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

Hence C is correct option.

Posted by

infoexpert22

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