Answer to Question #213339 in Differential Equations for anuj

Question

1.d²y/dx²-2dy/dx+5y=e^xtan2x

2.dy/dx+3y=3x²e^-3x

Answer 1

1.${d^2y\over dx^2}-2{dy\over dx}+5y=e^xtan2x$

The differential equation has the form

${d^2y\over dx^2}+p{dy\over dx}+qy=s$ where $p=-2$ , $q=5$ and $s=-e^xtan2x$

Solve corresponding homogenous linear equation

${d^2y\over dx^2}+p{dy\over dx}+qy=0$

Its characteristic equation is given as

$q+(k^2+kp)=0\implies5+(k^2-2k)=0 \implies k^2-2k+5=0$

$\therefore$ $k_1=1-2i$ and $k_2=1+2i$

$\implies$ $y=C_1e^{k_1x}+C_2e^{k_2x}=C_1e^{(1-2i)x}+C_2e^{(1+2i)x}$

Solve the inhomogeneous equation

${d^2y\over dx^2}+p{dy\over dx}+qy=s$

By use of variation of parameters method, suppose that $C_1$ and $C_2$ are functions of $x$

$\therefore$ The general solution is

$y=C_1(x)e^{(1-2i)x}+C_2(x)e^{(1+2i)x}$

To find $C_1(x)$ and $C_2 (x)$ ,by the method of variation of parameters ,we find the solution from the system:

$y_1(x){d\over dx}C_1(x)+y_2(x){d\over dx}C_2(x)=0$

${d\over dx}C_1(x){d\over dx}y_1(x)+{d\over dx}C_2(x){d\over dx}y_2(x)=f(x)$

Where $y_1(x)$ and $y_2(x)$ are linearly independent particular solutions of linear ordinary differential Equations

$y_1(x)=e^{x(1-2i)}$ , $C_1=1$ and $C_2=0$

$y_2(x)=e^{x(1+2i)}$ , $C_1=0$ and $C_2=1$

And the free term is given as

$f(x)=e^xtan(2x)$

So the system has the form:

$e^{x(1-2i)}{d\over dx}C_1(x)+e^{x(1+2i)}{d\over dx}C_2(x)=0$

${d\over dx}C_1(x){d\over dx}e^{x(1-2i)}+{d\over dx}C_2(x){d\over dx}e^{x(1+2i)}=e^xtan(2x)$

solving the system gives

${d\over dx}C_1(x)={ie^{2ixtan(2x)}\over 4} \implies C_1(x)=\int{ie^{2ixtan(2x)}\over 4}dx$

${d\over dx}C_2(x)={ie^{-2ixtan(2x)}\over 4}\implies C_2(x)=-\int {ie^{-2ixtan(2x)}\over 4}dx$

Evaluate $\int{ie^{2ixtan(2x)}\over 4}dx={log(e^{2ix}-i)\over 8}-{log(e^{2ix}+i)\over 8}-{ie^{2ix}\over 8}$

$-\int{ie^{-2ixtan(2x)}\over 4}dx={log(e^{2ix}-i)\over 8}-{log(e^{2ix}+i)\over 8}+{ie^{2ix}\over 8}$

Substitute $C_1(x)$ and $C_2(x)$ to

$y=C_1(x)e^{(1-2i)x}+C_2(x)e^{(1+2i)x}$

$\implies y=({log(e^{2ix}-i)\over 8}-{log(e^{2ix}+i)\over 8}-{ie^{2ix}\over 8})e^{(1-2i)x}+({log(e^{2ix}-i)\over 8}-{log(e^{2ix}+i)\over 8}+{ie^{2ix}\over 8})e^{(1+2i)x}$

$\therefore y=(C_2sin(2x)+(C_1+{log(sin(2x)-1)\over 8}-{log(sin(2x)+1)\over 8})cos(2x)e^x$

2.${dy\over dx}+3y=3x^2e^{-3x}$

This differential equation has the form

${dy\over dx}+p(x)y=Q(x)$ ​​

Solve the correspondent linear homogenous equation

${dy\over dx}+p(x)y=0$ where $p(x)=3$ and $Q(x)=3x^2e^{-3x}$

$\implies {dy\over dx}=-p(x)y \implies {dy\over y}=-p(x)dx \implies \int{1\over y}dy =-\int p(x)dx$

$\therefore log(|y|)=-\int p(x)dx \implies |y|=e^{-\int p(x)dx}$

$\implies y_1=e^{-\int p(x)dx}=e^{-\int 3dx}=e^{C_1-3x}$

and $y_2=-e^{-\int p(x)dx}=-e^{-\int 3dx}=-e^{C_2-3x}$

This leads to correspondent solution for any constant C not equal to zero

$y=Ce^{-3x}$

Now we should solve the inhomogeneous equation

${dy\over dx}+p(x)y=Q(x)$

Use variation of parameters method by considering C as a function of x

$y=C(x)e^{-3x}$

Apply it in the original equation using rules for product differentiation of composite functions derivative .We find that

${d\over dx}C(x)=Q(x)e^{\int p(x)dx}$

Let use Q(x) and p(x) for this equation, we get the first order differential equation for C(x)

${d\over dx}C(x)=3x^3 \implies C(x)=\int3x^2dx=x^3+c$

use C(x) at

$y=C(x)e^{-3x}$

$\therefore$ $y=(C_1+x^3)e^{-3x}$