Answer to Question #160353 in Differential Equations for Saif

Question

Solve the following equations using Wronskian’s method d³y/dx³+3d²y/dx²+3dy/dx+y=(x+40)e^-x

Answer 1

y"'+3y"+3y'+y=(x+40)e^-x

^{Let x=t}

Let the general solution be

$y(t) =c_1y_1(t) +c_2y_2(t)+ c_3y_3(t)+y_p(t)$

y₁, y_{2 ,} y_{3-are the solutions of the homogeneous equation. These solutions may be obtained by the characteristic equation}

$\lambda^3+3\lambda^2+3\lambda+1=0$

$(\lambda+1)^3=0$

$\lambda_1=\lambda_2=\lambda_3=-1$

$y_1=y_2=y_3=e^{-t}$

Wronskian test for independence.

Let us calculate the Wronskian:

$W=det\begin{pmatrix} y_1 & y_2 & y_3 \\ y_1' & y_2' & y_3' \\ y_1'' & y_2'' & y_3'' \end{pmatrix}=det\begin{pmatrix} e^{-t} & e^{-t} & e^{-t} \\ -e^{-t} & -e^{-t} & -e^{-t} \\ e^{-t} & e^{-t} & e^{-t} \end{pmatrix}=3e^{-t}(-e^{-2t}+e^{-2t})=0$

W=0, so y_1, y_2,y₃ are dependent. So

$y_p(t)=f(t)=(x+40)e^{-t}$

Answer: y(x)=3ce^-x+(x+40)e^-x=e^-x(3c+x+40)