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Popular Calculus >

y^{''}+4y=sin(x)*sin(2x)

Solution

y^{^{''}} + 4 y = sin (x) \cdot sin (2 x)

Solution

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{2} cos (2 x) (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x) - sin^{2} (2 x) cos (x)) + \frac{1}{8} sin (2 x) (- \frac{1}{5} sin (5 x) + \frac{1}{3} sin (3 x))

Solution steps

y^{''} (x) + 4 y = sin (x) \cdot sin (2 x)

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{2} cos (2 x) (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x) - sin^{2} (2 x) cos (x)) + \frac{1}{8} sin (2 x) (- \frac{1}{5} sin (5 x) + \frac{1}{3} sin (3 x))

y = c_{1} cos (2 x) + c_{2} sin (2 x) - \frac{1}{2} cos (2 x) (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x) - sin^{2} (2 x) cos (x)) + \frac{1}{8} sin (2 x) (- \frac{1}{5} sin (5 x) + \frac{1}{3} sin (3 x))

Graph

Popular Examples

y^{''}+y^'=cos(x),y(0)=0,y^'(0)=1

4y^{''}-y=e^{(x/2)}+5

x^{''}+64x=sin(5t)

0.5y^{''}+50y=cos(t)

x^{''}+x=1,x(0)=2,x^'(0)=-1

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+4y=sin(x)*sin(2x) ?
The solution for y^{''}+4y=sin(x)*sin(2x) is