Question

Solve the following differential equation:

1. (D²+D)y=sin x

2. (D²+4D+5) y= 50x + 13e^3x

3. (D³+D²-4D-4)y= 4 sin x

4.(D³-D)y=x

5. (D²-4D+4)y=e^x

Answer 1

1.

$k^2+1=0$

$k=\pm i$

$y_h=c_1cosx+c_2sinx$

$y_p=Axcosx+Bxsinx$

$y'_p=Acosx-Axsinx+Bsinx+Bxcosx=(A+Bx)cosx+(B-Ax)sinx$

$y''_p=Bcosx-(A+Bx)sinx+(B-Ax)cosx-Asinx$

$Bcosx-(A+Bx)sinx+(B-Ax)cosx-Asinx+Axcosx+Bxsinx=sinx$

$B=0$

$-A-A=1$

$A=-1/2$

$y=y_h+y_p=c_1cosx+c_2sinx-xcosx/2$

2.

$k^2+4k+5=0$

$k=\frac{-4\pm\sqrt{16-20}}{2}=-2\pm 2i$

$y_h=e^{-2x}(c_1cos2x+c_2sin2x)$

$y_{p1}=Ax+B$

$4A+5Ax+5B=50x$

$A=10,B=-40/5=-8$

$y_{p1}=10x-8$

$y_{p2}=Ae^{3x}$

$9Ae^{3x}+12e^{3x}+5e^{3x}=13e^{3x}$

$A=-4/9$

$y_{p2}=-4e^{3x}/9$

$y=e^{-2x}(c_1cos2x+c_2sin2x)+10x-8-4e^{3x}/9$

3.

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

$k(k^2-4)+k^2-4=0$

$k_1=-1,k_2=-2,k_3=2$

$y_h=c_1e^{-x}+c_1e^{-2x}+c_1e^{2x}$

$y_p=Acosx+Bsinx$

$y'_p=Bcosx-Asinx$

$y''_p=-Bsinx-Acosx$

$y'''_p=-Bcosx+Asinx$

$-Bcosx+Asinx -Bsinx-Acosx-4(Bcosx-Asinx)-4(Acosx+Bsinx)=$

$=4 sin x$

$-5B-5A=0$

$5A-5B=4$

$10A=4\implies A=2/5,B=-2/5$

$y=c_1e^{-x}+c_1e^{-2x}+c_1e^{2x}+2(cosx-sinx)/5$

4.

$k^3-1=0$

$(k-1)(k^2+k+1)=0$

$k_1=1$

$k^2+k+1=0$

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

$y_h=c_1e^x+e^{-x/2}(c_2cos(x\sqrt 3/2)+c_3sin(x\sqrt 3/2))$

$y_p=Ax+B$

$-Ax-B=x$

$A=-1,B=0$

$y=c_1e^x+e^{-x/2}(c_2cos(x\sqrt 3/2)+c_3sin(x\sqrt 3/2))-x$

5.

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

$k_{1,2}=2$

$y_h=c_1e^{2x}+c_2xe^{2x}$

$y_p=Ae^x$

$A-4A+4A=1$

$A=1$

$y=c_1e^{2x}+c_2xe^{2x}+e^x$