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Popular >

y^{''}-10y^'+9y=5t,y(0)=-1,y^'(0)=2

Solution

y^{^{''}} - 10 y^{^{'}} + 9 y = 5 t, y (0) = - 1, y^{^{'}} (0) = 2

Solution

y = (2 - \frac{131}{81}) e^{9 t} - 2 e^{t} + \frac{5 t}{9} + \frac{50}{81}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 9 y = 5 t

Solve linear ODE:

y = (2 - \frac{131}{81}) e^{9 t} - 2 e^{t} + \frac{5 t}{9} + \frac{50}{81}

y = (2 - \frac{131}{81}) e^{9 t} - 2 e^{t} + \frac{5 t}{9} + \frac{50}{81}

Popular Examples

y^{''}-2y^'+2y=(e^x)/(sin(x))

y^{^{''}} - 2 y^{^{'}} + 2 y = \frac{e ^{x}}{sin ( x )}

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

y^{''}+4y^'+4y=8e^{-2x}

y^{^{''}} + 4 y^{^{'}} + 4 y = 8 e^{- 2 x}

y^{''}-8y^'+16y=e^{4x}ln(9+x^2)

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x} ln (9 + x^{2})

(d^2y)/(dx^2)-8(dy)/(dx)+16y=e^{4x}

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 16 y = e^{4 x}

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-10y^'+9y=5t,y(0)=-1,y^'(0)=2 ?
The solution for y^{''}-10y^'+9y=5t,y(0)=-1,y^'(0)=2 is y=(2-131/81)e^{9t}-2e^t+(5t)/9+50/81