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Popular >

y^'=2y+xe^x

Solution

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

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Popular Examples

limit as x approaches+0 of (ln(x))/x

x \to + 0 lim (\frac{ln ( x )}{x})

tangent of y=(1+4x)^8,(0,1)tangent of

y = (1 + 4 x)^{8}, (0, 1)

limit as x approaches 0+of inf(t,y,x)inity

x \to 0 + lim (in f (t, y, x) ini t y)

(dy)/(dx)+3y=3x^2e^{-3x}

\frac{d y}{d x} + 3 y = 3 x^{2} e^{- 3 x}

integral of (18)/(9x^2-x^4)

\int \frac{18}{9 x ^{2} - x ^{4}} d x

Frequently Asked Questions (FAQ)

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