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Popular >

y^'+y-e^x=0

Solution

y^{^{'}} + y - e^{x} = 0

Solution

y = \frac{e ^{x}}{2} + \frac{c _{1}}{e ^{x}}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{x}}{2} + \frac{c _{1}}{e ^{x}}

y = \frac{e ^{x}}{2} + \frac{c _{1}}{e ^{x}}

Popular Examples

area y=x,y=x^3,[-1,2]area

y = x, y = x^{3}, [- 1, 2]

integral of-e^{2t}

\int - e^{2 t} d t

((3y^2-t^2))/(y^5)(dy)/(dt)+t/(2y^4)=0

\frac{( 3 y ^{2} - t ^{2} )}{y ^{5}} \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

(\partial)/(\partial y)(cos(x^3y^3))

\frac{\partial}{\partial y} (cos (x^{3} y^{3}))

(\partial)/(\partial x)(-2xysin(xy^2))

\frac{\partial}{\partial x} (- 2 x y sin (x y^{2}))

Frequently Asked Questions (FAQ)

What is the solution for y^'+y-e^x=0 ?
The solution for y^'+y-e^x=0 is y=(e^x)/2+(c_{1})/(e^x)