Suppose that a fourth order differential equation has a solution

Find such a differential equation, assuming it is homogeneous and has constant coefficients.

Find the general solution to this differential equation. In your answer, use

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y=-7 e^{2 x} x \sin xNeglect the constant -7 at the moment.We have e^{2 x} x \sin xWhen we get complex roots in homogeneous differential equation with constant coefficients, then solution is of the form y=e^{a x} \cos b x+e^{a x} \sin b x where a is real part of complex root and b is imaginary part of complex root.Here we have e^{2 x} \sin x, i.e. a=2 and b=1.i.e. we have m=2+i and m=2-i as two pair of complex roots. \bigcirc R, we get\begin{array}{ll}m-(2+i)=0 \text { and } & m-(2-i)=0 \\m-2-i=0 \quad \text { and } & m-2+i=0\end{array}combine both, we get\begin{array}{l}(m-2-i)(m-2+i)=0 \\(m-2)^{2}-(i)^{2}=0 \\m^{2}-4 m+4-i^{2}=0 \\m^{2}-4 m+4-(-1)=0 \\m^{2}-4 m+5=0\end{array}Also, remember that when roots are repeated, then we multiply solution with x, Here, we are given e^{2 x} x \sin x, i.e. Above complex root is repeated twice.Therefore, we get\left(m^{2}-4 m+5\right)\left(m^{2}-4 m+5\right)=0Upon simplifying, we getm^{4}-8 m^{3}+26 m^{2}-40 m+25=0Replace m with differentail operator D and multiply by m, we get\begin{array}{l}\left(D^{4}-8 D^{3}+26 D^{2}-40 D+25\right) y=0 \\\frac{d^{4} y}{d x^{4}}-8 \frac{d^{3} y}{d x^{3}}+26 \frac{d^{2} y}{d x^{2}}-40 \frac{d y}{d x}+25 y=0\end{array}is the required differential equation.General Solution of above differential equation isy=A e^{2 x} \sin x+B e^{2 x} \cos x+C x e^{2 x} \sin x+D x e^{2 x} \cos x ...