Answer to Question #122613 in Differential Equations for jessica

Question

Solve the differential equation by variation of parameters.
4y'' − 8y' + 8y = e^x sec x

y(x) =______

Answer 1

The given equation can be written as:

(4D²-8D+8)y = e^xsecx

complementary equation is:

4D²-8D+8=0

D=4+4i,4-4i

Since they are complex roots, general solution is

y= e^4x(acos4x + bsin4x)

Particular solution by variation of parameters:

Let z=e^4xcos4x and u= e^4xsin4x

Wronskian(z,u) =$\begin{vmatrix} e^{4x}cos4x & e^{4x}sin4x \\ 4e^{4x}cos4x - 4e^{4x}sin4x & 4e^{4x}sin4x + 4e^{4x}cos4x \end{vmatrix}$

=4e^4x(sin²4x+cos²4x)=4e^4x

Particular solution is

=-z($\int (u/W) dt+u(\int(z/W)dt$

= -e^4xcos4x ($\int (e^{4x}sin4x/4e^{4x})$ $+e^{4x}sin4x (\int(e^{4x}cos4x /4e^{4x})$

=$-e^{4x}cos4x (-cos4x/16) + e^{4x}sin4x (sin4x/16)+c$

So the solution is

y=

e^4x(acos4x + bsin4x)+ $-e^{4x}cos4x (-cos4x/16) + e^{4x}sin4x (sin4x/16)+c$