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Popular >

y^'=2y+xe^x,y(0)=-1

Solution

y^{^{'}} = 2 y + x e^{x}, y (0) = - 1

Solution

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

Solution steps

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

Solve linear ODE:

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

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

Popular Examples

derivative of (55)/(0.01x^2+1)derivative of

\frac{55}{0.01 x ^{2} + 1}

integral of tan^4(4t)

\int tan^{4} (4 t) d t

(\partial)/(\partial y)(xy+ln(x+y))

\frac{\partial}{\partial y} (x y + ln (x + y))

normal of y=2xe^x,(0,0)normal of

y = 2 x e^{x}, (0, 0)

maclaurin cos(pix)maclaurin

cos (π x)

Frequently Asked Questions (FAQ)

What is the solution for y^'=2y+xe^x,y(0)=-1 ?
The solution for y^'=2y+xe^x,y(0)=-1 is y=-e^xx-e^x