Answer to Question #221499 in Differential Equations for Anuj

Question

By method variation of parameters solve the ODE(D^2+4)y=2 cosec2x?

Answer 1

Solution;

The general solution of the equation is given by;

y=y_p+y_h

The equation has an auxiliary equation of the form;

P(m)=m²+4=0

m²=-4

m=-2i and m=2i

Hence the homogenous solutions is;

y_h=C₁cos(2x)+C₂sin(2x)

Here;

y₁=cos(2x)

y₂=sin(2x)

X=2cosec(2x)

$W=\begin{vmatrix} cos(2x) & sin(2x)\\ -2sin(2x) & 2cos(2x) \end{vmatrix}$

det(W)=2(cos²(2x)+sin²(2x))=2(1)=2$\ne0$

For the particular solution;

Let y_p=uy₁+vy₂

$u=-\int\frac{y_2X}{W}dx$

$u=-\int\frac{sin(2x)2cosec(2x)}{2}$ =$-\int1dx$

u=-x

$v=\int\frac{y_1X}{W}dx$

$v=\int\frac{cos(2x)2cosec(2x)}{2}dx$ =$\int\frac{cos(2x)}{sin(2x)}dx$ =$\frac12ln(sin(2x))$

Hence;

y_p=-xcos(2x)+$\frac12$ln(sin(2x))sin(2x)

Hence the required solution is ;

y=C₁cos(2x)+C₂sin(2x)-xcos(2x)+$\frac12$sin(2x)ln(sin(2x))