Get Answers to all your Questions

header-bg qa

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)

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

Posted by

infoexpert22

View full answer