Question

Solve the Differential equation

(D^2+2D+1)y=xe^-x. cos2x​

Answer 1

This is the solution of your problem.

Answer 2

Answer:

Concept:

Solving the Differential equation.

Step-by-step explanation:

Given:

$&\left(D^{2}+2 D+1\right) y=x e^{-x} \cos2 x$

Find:

Solve the differential equation $&\left(D^{2}+2 D+1\right) y=x e^{-x} \cos2 x$  

Solution:

$&m^{2}+2 m+1=0\\&m^{2}+m+m+1=0\\&m(m+1)+1(m+1)=0\\&(m+1)(m+1)=0$

m = -1, -1

$&C_{} F=\left(x c_{1}+c_{2}\right) e^{m_{1} x}\\&C_{} F=\left(x C_{1}+C_{2}\right) e^{-x}\\\\&P_{} I=\frac{k e^{-x} \cos 2x}{\left(D^{2}+2 D+1\right)}$

$&P_{} I=\frac{e^{-x}(x \cos x)}{D^{2}+2 D+1}\\&\text { By applying, } \frac{1}{f(D)} e^{a x} V=e^{a x}\left(\frac{1}{f(D+a)} V\right)$

$&P_{} I=e^{-x} \frac{(x \cos 2x)}{(D-1)^{2}+2(D-1)+1}=e^{-x} \frac{(x \cos 2x)}{D^{2}+1-2 D+3 D-2+1}\\\\&P_{} I=e^{-x} \frac{(x \cos 2x)}{D^{2}+2 x}=e^{-x} \frac{(x \cos 2x)}{D^{2}}\\\\&\text { PI }=e^{-x} \frac{1}{D} \int \frac{x}{\text { (I) }} \cos 2x d x$

$&P_{} I=e^{-x} \frac{1}{D}\left[x \sin x-\int \sin x d x\right]$

$&P_{} I=e^{-x} \frac{I}{D}[x \sin x+\cos x]$

$&P_{I}=e^{-x} \int(x \sin x+\cos x) d x \\\\&P_{} I=e^{-x}\left[-x \cos x+\int \cos x d x+\int \cos x d x\right]$

$&P_{} I=e^{-x}[-x \cos x+2 \sin x] \\&\left.P_{} I=e^{-x}[2 \sin x-x \cos x]\right)$

This is the required equation.

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