Answer to Question #218267 in Differential Equations for Neha

Question

Solve (d^2y)/(dx^2)+y=sec^2x

Answer 1

${d^2y\over dx^2}+y=sec^2(x)$

But $sec^2(x)=tan^2(x)+1$

Rewriting the differential equation, we have

${d^2y\over dx^2}+y=tan^2(x)+1......................(i)$

The corresponding homogenous linear equation is

${d^2y\over dx^2}+y=0........................(ii)$

To solve this homogenous linear equation $(ii)$ ,Find the roots of characteristic equation

$k^2+1=0 \implies$ $k_1=-i$ or $k_2=i$

The two roots of the characteristic equation are purely imaginary

$\therefore y(x)=C_1sin(x|k_1|)+C_2cos(x|k_2|)$

$\implies y(x)=C_1sin(x)+C_2cos(x)...................(iii)$

Equation $(iii)$ is our general solution

Now we should solve the inhomogeneous equation $(i)$ Using variation of parameters method

Suppose $C_1$ is function of $x$ and $C_2$ is also a function of $x$

The general solution is:

$\implies y(x)=C_1(x)sin(x)+C_2(x)cos(x).................(iv)$

Where $C_1(x)$ and $C_2(x)$ by 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 solutions of linear Ordinary differential equations and $f(x)$ is the free term and are given as

$y_1(x)=sin(x)$ , $y_2(x)=cos(x)$ and $f(x)=tan^2(x)+1$

Substituting to the system above, we have

$sin(x){d\over dx}C_1(x)+cos(x){d\over dx}C_2(x)=0$

${d\over dx}C_1(x){d\over dx}sin(x)+{d\over dx}C_2(x){d\over dx}cos(x)=tan^2(x)+1$

But ${d\over dx}sin(x)=cos (x)$ and ${d\over dx}cos(x)=-sin(x)$

Substituting to our system, we get:

$sin(x){d\over dx}C_1(x)+cos(x){d\over dx}C_2(x)=0$

$cos(x){d\over dx}C_1(x)-sin(x){d\over dx}C_2(x)=tan^2(x)+1$

Solving the system simultaneously, We get

${d\over dx}C_1(x)={1\over cos(x)}$ and ${d\over dx}C_2(x)=-{sin(x)\over cos^2(x)}$

let's solve ${d\over dx}C_1(x)={1\over cos(x)}$ and ${d\over dx}C_2(x)=-{sin(x)\over cos^2(x)}$

${d\over dx}C_1(x)={1\over cos(x)} \implies dC_1(x)={dx\over cos(x)}$

Integrating both sides we have

$\int dC_1(x)=\int {dx\over cos(x)}\implies C_1(x)=C_3-{log(sin(x)-1)\over 2}+{log(sin(x)+1)\over 2}$

${d\over dx}C_2(x)=-{sin(x)\over cos^2(x)} \implies dC_2(x)=-{sin(x)\over cos^2(x)}dx$

Integrating both sides we have

$\int dC_2(x)=-\int{sin(x)\over cos^2(x)}dx\implies C_2(x)=C_4-{1\over cos(x)}$

Now, substitute $C_1(x)$ and $C_2(x)$ to the general solution $(iv)$

$y(x)=C_1(x)sin(x)+C_2(x)cos(x)$

$=(C_3-{log(sin(x)-1)\over 2}+{log(sin(x)+1)\over 2})sin(x)+(C_4-{1\over cos(x)})(x)cos(x)$

$\therefore y(x)=C_2cos(x)+(C_1-{log(sin(x)-1)\over 2}+{log(sin(x)+1)\over 2})sin(x)-1$