Question

solve the following differential equation(D^2+2D+1)y= xcosx​

Answer 1

Answer:

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$solve \: the \: following \: differential \: equation$

$( {d}^{2} + 2d + 1)y = xcosx$

$ie \: {(d + 1)}^{2}y = xcosx$

$differentitating \: both \: side \: with \: respect \: to \: x$

$( {d + 1)}^{2}y = xcosx$

$= {(d + 1)}^{2} \times \frac{dy}{dx} + 2(d + 1)y = x \times - sinx + cosx$

${(d + 1)}^{2} \frac{dy}{dx} + 2(d + 1)y = - xsinxcosx$

$( {d}^{2} + 2d + 1) \frac{dy}{dx} + (2d + 2)y + xsinxcosx = 0$

$( {d}^{2} + 2d + 1) \frac{dy}{dx} = - (2d + 2) - xsinxcosx$

$\frac{dy}{dx} = \frac{ - (2d + 2) - xsinxcosx}{ {d}^{2} + 2d + 1 }$

$\frac{dy}{dx} = \frac{ - 2(d + 1) - xcosxsinx}{ {(d + 1)}^{2} }$

$\frac{dy}{dx} = - \frac{2 - xsinxcosx}{d + 1}$