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Popular >

y^{''}-y^'=2e^x

Solution

y^{^{''}} - y^{^{'}} = 2 e^{x}

Solution

y = c_{1} e^{x} + c_{2} + 2 x e^{x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} + 2 x e^{x}

y = c_{1} e^{x} + c_{2} + 2 x e^{x}

Graph

Popular Examples

5y^{''}+25y^'+1/(10^{-2)}y=10e^{-0/4}

5 y^{^{''}} + 25 y^{^{'}} + \frac{1}{1 0 ^{- 2}} y = 10 e^{- \frac{0}{4}}

y^{''}+2y^'=2x+4

y^{^{''}} + 2 y^{^{'}} = 2 x + 4

y^{''}+4y^'+3y=e^t+sin(t)

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{t} + sin (t)

y^{''}-2y^'-3y=2e^{3t},y(0)=1,y^'(0)=0

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e^{3 t}, y (0) = 1, y^{^{'}} (0) = 0

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

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

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-y^'=2e^x ?
The solution for y^{''}-y^'=2e^x is y=c_{1}e^x+c_{2}+2xe^x