Answer to Question #200436 in Differential Equations for Ayush

Question

Using the method of undetermined coefficients, find the general solution of the DE y^iv -2y^m+2yⁿ=3e^-x+2e^-x+e^-xsin x.

Answer 1

$y^{(iv)} -2y'''+2y''=3e^{-x}+2e^{-x}+e^{-x}sin x=5e^{-x}+e^{-x}sin x$

characteristic equation:

$k^4-2k^3+2k^2=0$

$k^2(k^2-2k+2)=0$

$k_{1,2}=0$

$k_{3,4}=\frac{2\pm \sqrt{4-8}}{2}=1\pm i$

$y_c=c_1+c_2x+e^x(c_3cosx+c_4sinx)$

$y_{p1}=Ae^{-x}$

$Ae^{-x}+2Ae^{-x}+2Ae^{-x}=5e^{-x}$

$A=1$

$y_{p1}=e^{-x}$

$y_{p2}=e^{-x}(Acosx+Bsinx)$

$y'_{p2}=-e^{-x}(Acosx+Bsinx)+e^{-x}(-Asinx+Bcosx)=$

$=e^{-x}((B-A)cosx-(A+B)sinx)$

$y''_{p2}=-e^{-x}((B-A)cosx-(A+B)sinx)+e^{-x}((A-B)sinx-(A+B)cosx)=$

$=e^{-x}(-2Bcosx+2Asinx)$

$y'''_{p2}=-2e^{-x}(Asinx-Bcosx)+2e^{-x}(Acosx+Bsinx)=$

$=2e^{-x}((A+B)cosx+(B-A)sinx)$

$y^{(iv)}_{p2}=-2e^{-x}((A+B)cosx+(B-A)sinx)+$

$+2e^{-x}(-(A+B)sinx+(B-A)cosx)=-4e^{-x}(Acosx+Bsinx)$

$-4e^{-x}(Acosx+Bsinx)-4e^{-x}((A+B)cosx+(B-A)sinx)+$

$+2e^{-x}(-2Bcosx+2Asinx)=e^{-x}sin x$

$-4A-4(A+B)-4B=0\implies A+B=0$

$-4B-4(B-A)+4A=1\implies A-B=1$

$A=1/2,B=-1/2$

$y_{p2}=e^{-x}(cosx-sinx)/2$

$y=y_c+y_{p1}+y_{p2}=$

$=c_1+c_2x+e^x(c_3cosx+c_4sinx)+e^{-x}+e^{-x}(cosx-sinx)/2$