Answer to Question #283046 in Differential Equations for Varun

Question

solve the differential equation by the method of variation of parameters d²y/dx²+9y=sec3x

Answer 1

The given differential equation can be written as;

$(D^2+9)y=sec3x , where\ D=\frac{d}{dx}$

Auxiliary equation is $(D^2+9)=0$

$\implies D=\pm3i$

Hence, C.F is,

$y_c=C_1\ cos3x+C_2\ sin3x$

Now we take $y_1=cos 3x, y_2=sin3x$

$w=\begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}=\begin{vmatrix} cos3x & sin3x \\ -3sin3x & 3cos3x \end{vmatrix}=3$

Particular integral is,

$y_p=-y_1\int \frac{y_2X}{w}dx+y_2\int \frac{y_1X}{w}dx$

$=-cos3x\int \frac{sin3xsec3x}{3}dx+sin3x\int \frac{cos3xsec3x}{3}dx$

$=-\frac{cos3x}{3}\int tan3xdx+\frac{sin3x}{3}\int dx$

$=\frac{cos 3x\ log(cos3x)}{9}+\frac{x\ sin3x}{3}$

Hence, the general solution is;

y(x)=$y_c+y_p$

$y(x)=C_1\ cos3x+C_2\ sin3x+\frac{cos 3x\ log(cos3x)}{9}+\frac{x\ sin3x}{3}$