Answer to Question #302574 in Differential Equations for haru

Question

Solve the following initial value problems using method of undetermined coefficients: (ii) y''+y' = 5xsinx, y(0) = 0,y(0) = 1.

Answer 1

$y''+y'=5xsinx\\ y(0)=0, y'(0)=1\\ y=y_1+y_2\\ 1) y''+y'=0\\ \lambda^2+\lambda=0\\ \lambda(\lambda+1)=0\\ \lambda_1=0, \lambda_2=-1\\ y_1=C_1+C_2e^{-x}$

$2) y_2=(ax+b)cosx+(kx+p)sinx\\ y_2'=acosx-(ax+b)sinx+\\+ksinx+(kx+p)cosx=\\ =(kx+p+a)cosx+(-ax-b+k)sinx\\ y_2''=kcosx-(kx+p+a)sinx-\\-asinx+(-ax-b+k)cosx=\\ =(-ax-b+2k)cosx+(-kx-p-2a)sinx$

$(-ax-b+2k)cosx+(-kx-p-2a)sinx+\\ +(kx+p+a)cosx+(-ax-b+k)sinx=\\=5xsinx$

$xcosx: -a+k=0\\ cosx: -b+2k+p+a=0\\ xsinx:-k-a=5\\ sinx:-p-2a-b+k=0$

$a=-\frac{5}{2}, k=-\frac{5}{2}, b=\frac{5}{2}, p=5\\ y_2=(-\frac{5}{2}x+\frac{5}{2})cosx+(-\frac{5}{2}x+5)sinx$

$y=C_1+C_2e^{-x}+\\+(-\frac{5}{2}x+\frac{5}{2})cosx+(-\frac{5}{2}x+5)sinx\\ y'=-C_2e^{-x}-(-\frac{5}{2}x+\frac{5}{2})sinx-\\ -\frac{5}{2}cosx+(-\frac{5}{2}x+5)cosx-\frac{5}{2}sinx\\ y(0)=0\\ C_1+C_2+\frac{5}{2}=0\\ y'(0)=1\\ -C_2-\frac{5}{2}+5=1\\ C_2=\frac{3}{2}, C_1=-4$

$y=-4+\frac{3}{2}e^{-x}+\\+(-\frac{5}{2}x+\frac{5}{2})cosx+(-\frac{5}{2}x+5)sinx$