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y^{''}-4y^'-4t+4=0,y(0)=2,y^'(0)=9

Solution

y^{^{''}} - 4 y^{^{'}} - 4 t + 4 = 0, y (0) = 2, y^{^{'}} (0) = 9

Solution

y = (2 + \frac{1}{16}) e^{4 t} - \frac{1}{16} - \frac{t ^{2}}{2} + \frac{3 t}{4}

Solution steps

y^{''} (t) - 4 y^{'} (t) - 4 t + 4 = 0

Solve linear ODE:

y = (2 + \frac{1}{16}) e^{4 t} - \frac{1}{16} - \frac{t ^{2}}{2} + \frac{3 t}{4}

y = (2 + \frac{1}{16}) e^{4 t} - \frac{1}{16} - \frac{t ^{2}}{2} + \frac{3 t}{4}

Graph

Popular Examples

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2(d^2y)/(dx^2)+4(dy)/(dx)-6y=14e^x

2 \frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 6 y = 14 e^{x}

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\frac{1}{4} y^{^{''}} + 2 y^{^{'}} + 40 y = - \frac{20}{3} t - \frac{4}{3}

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

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

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-4y^'-4t+4=0,y(0)=2,y^'(0)=9 ?
The solution for y^{''}-4y^'-4t+4=0,y(0)=2,y^'(0)=9 is y=(2+1/16)e^{4t}-1/16-(t^2)/2+(3t)/4