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Popular >

y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=1

Solution

y^{^{''}} + 4 y = t^{2} + 10 e^{t}, y (0) = 0, y^{^{'}} (0) = 1

Solution

y = - \frac{15}{8} cos (2 t) - \frac{1}{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + 2 e^{t}

Solution steps

y^{''} (t) + 4 y = t^{2} + 10 e^{t}

Solve linear ODE:

y = - \frac{15}{8} cos (2 t) - \frac{1}{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + 2 e^{t}

y = - \frac{15}{8} cos (2 t) - \frac{1}{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8} + 2 e^{t}

Graph

Popular Examples

y^{'''}+3y^{''}+3y^'+y=x^{-3}e^{-x}

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = x^{- 3} e^{- x}

y^{''}-y^'+12y=t,y(0)=1,y^'(0)=1

y^{^{''}} - y^{^{'}} + 12 y = t, y (0) = 1, y^{^{'}} (0) = 1

25y^{''}-y=xe^{1/5 x}

25 y^{^{''}} - y = x e^{\frac{1}{5} x}

3x^{''}+10x=2sin(3t)

3 x^{^{''}} + 10 x = 2 sin (3 t)

y^{''}-y^'-6y=(10x+7)e^{3x}

y^{^{''}} - y^{^{'}} - 6 y = (10 x + 7) e^{3 x}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+4y=t^2+10e^t,y(0)=0,y^'(0)=1 is y=-15/8 cos(2t)-1/2 sin(2t)+(t^2)/4-1/8+2e^t