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Popular >

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

Solution

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

y^{^{''}} + 2 y^{^{'}} + 2 y = a cos (x)

y^{''}-2y^'+10y=40,y(0)=0,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} + 10 y = 40, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 4 \cdot y = - 3

u^{''}-a*\H(x)=x

u^{^{''}} - a \cdot H (x) = x

x^{''}+5x^'+4x=t+1

x^{^{''}} + 5 x^{^{'}} + 4 x = t + 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