Answer to Question #218096 in Differential Equations for Syed Hassan abdull

Question

Find the general solution of the differential equation

(D^2+1)y=x^3+e^xcosx

Answer 1

$\left( {{D^2} + 1} \right)y = y'' + y = {x^3} + {e^x}\cos x$

Сharacteristic equation:

${k^2} + 1 = 0$

${k^2} = - 1$

$k = \pm i$

Then the general solution of the homogeneous equation is

${y_0} = {C_1}\cos x + {C_2}\sin x$

We will seek a particular solution in the form

$\widetilde y = A{x^3} + B{x^2} + Cx + D + {e^x}\left( {E\cos x + F\sin x} \right) \Rightarrow$

$\Rightarrow {\widetilde y^\prime } = 3A{x^2} + 2Bx + C + {e^x}\left( {E\cos x + F\sin x} \right) + {e^x}\left( { - E\sin x + F\cos x} \right) \Rightarrow$

$\Rightarrow {\widetilde y^{\prime \prime }} = 6Ax + 2B + {e^x}\left( {E\cos x + F\sin x} \right) + {e^x}\left( { - E\sin x + F\cos x} \right) + {e^x}\left( { - E\sin x + F\cos x} \right) + {e^x}\left( { - E\cos x - F\sin x} \right)$

Substitute the obtained values ​​into the original equation:

$6Ax + 2B + {e^x}\left( {E\cos x + F\sin x} \right) + {e^x}\left( { - E\sin x + F\cos x} \right) + {e^x}\left( { - E\sin x + F\cos x} \right) + {e^x}\left( { - E\cos x - F\sin x} \right) + A{x^3} + B{x^2} + Cx + D + {e^x}\left( {E\cos x + F\sin x} \right) = {x^3} + {e^x}\cos x$

$A{x^3} + B{x^2} + x\left( {6A + C} \right) + D + 2B + {e^x}\cos x(E + F + F - E + E) + {e^x}\sin x(F - E - E - F + F) = {x^3} + {e^x}\cos x$

$\left\{ \begin{matrix} A = 1\\ B = 0\\ 6A + C = 0\\ D + 2B = 0\\ 2F + E = 1\\ - 2E + F = 0 \end{matrix} \right. \Rightarrow A = 1,\,B = 0,\,C = - 6,\,D = 0,\,F = \frac{2}{5},\,E = \frac{1}{5}$

So,

$\widetilde y = {x^3} - 6x + {e^x}\left( {\frac{1}{5}\cos x + \frac{2}{5}\sin x} \right)$

$y = {y_0} + \widetilde y = {C_1}\cos x + {C_2}\sin x + {x^3} - 6x + {e^x}\left( {\frac{1}{5}\cos x + \frac{2}{5}\sin x} \right)$

Answer: $y = {C_1}\cos x + {C_2}\sin x + {x^3} - 6x + {e^x}\left( {\frac{1}{5}\cos x + \frac{2}{5}\sin x} \right)$