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Popular Calculus >

(d^3+4d)y=sin^2(x)

Solution

(d^{3} + 4 d) y = sin^{2} (x)

Solution

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{8} (x - \frac{1}{2} sin (2 x)) - \frac{1}{4} cos (2 x) (\frac{1}{2} sin^{2} (x) sin (2 x) - \frac{1}{4} (x - \frac{1}{4} sin (4 x))) - \frac{1}{8} sin^{4} (x) sin (2 x)

Solution steps

(d^{3} + 4 d) y = sin^{2} (x)

Solve linear ODE:

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{8} (x - \frac{1}{2} sin (2 x)) - \frac{1}{4} cos (2 x) (\frac{1}{2} sin^{2} (x) sin (2 x) - \frac{1}{4} (x - \frac{1}{4} sin (4 x))) - \frac{1}{8} sin^{4} (x) sin (2 x)

y = c_{1} + c_{2} cos (2 x) + c_{3} sin (2 x) + \frac{1}{8} (x - \frac{1}{2} sin (2 x)) - \frac{1}{4} cos (2 x) (\frac{1}{2} sin^{2} (x) sin (2 x) - \frac{1}{4} (x - \frac{1}{4} sin (4 x))) - \frac{1}{8} sin^{4} (x) sin (2 x)

Graph

Popular Examples

y^{''}+y^'=t^2+2t

y^{^{''}} + y^{^{'}} = t^{2} + 2 t

y^{''}+9y=10,e^{-t}y(0)y^'(0)=0

y^{^{''}} + 9 y = 10, e^{- t} y (0) y^{^{'}} (0) = 0

y^{''}-4y^'+3y=xcos(x)

y^{^{''}} - 4 y^{^{'}} + 3 y = x cos (x)

y^{''}+y=(cos(x)+1)/(sin(x))

y^{^{''}} + y = \frac{cos ( x ) + 1}{sin ( x )}

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

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

Frequently Asked Questions (FAQ)

What is the solution for (d^3+4d)y=sin^2(x) ?
The solution for (d^3+4d)y=sin^2(x) is