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Popular >

y^{''}+4y=tsin^2(t)

Solution

y^{^{''}} + 4 y = t sin^{2} (t)

Solution

y = c_{1} cos (2 t) + c_{2} sin (2 t) + (- \frac{1}{4} t sin^{4} (t) - \frac{1}{16} sin^{3} (t) cos (t) + \frac{3}{32} t - \frac{3}{64} sin (2 t)) cos (2 t) + \frac{1}{128} (- 8 t^{2} + 32 t sin^{2} (t) sin (2 t) + 4 t sin (4 t) - 16 sin^{4} (t) + cos (4 t)) sin (2 t)

Solution steps

y^{''} (t) + 4 y = t sin^{2} (t)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + (- \frac{1}{4} t sin^{4} (t) - \frac{1}{16} sin^{3} (t) cos (t) + \frac{3}{32} t - \frac{3}{64} sin (2 t)) cos (2 t) + \frac{1}{128} (- 8 t^{2} + 32 t sin^{2} (t) sin (2 t) + 4 t sin (4 t) - 16 sin^{4} (t) + cos (4 t)) sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + (- \frac{1}{4} t sin^{4} (t) - \frac{1}{16} sin^{3} (t) cos (t) + \frac{3}{32} t - \frac{3}{64} sin (2 t)) cos (2 t) + \frac{1}{128} (- 8 t^{2} + 32 t sin^{2} (t) sin (2 t) + 4 t sin (4 t) - 16 sin^{4} (t) + cos (4 t)) sin (2 t)

Graph

Popular Examples

y^{''}-2y^'-15=0

y^{^{''}} - 2 y^{^{'}} - 15 = 0

(D^2-3D+2)y=xe^x

(D^{2} - 3 D + 2) y = x e^{x}

y^{''}+y=8sin(x)+4cos(x)+1

y^{^{''}} + y = 8 sin (x) + 4 cos (x) + 1

x^{''}+4x^'+4=0

x^{^{''}} + 4 x^{^{'}} + 4 = 0

y^{''}-2y^'+y=(e^t)/((1+t^2)+3e^t)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{( 1 + t ^{2} ) + 3 e ^{t}}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+4y=tsin^2(t) ?
The solution for y^{''}+4y=tsin^2(t) is