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Popular >

xy^'+(1+x)=e^{-x}

Solution

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

Solution

y = Ei (- x) - x - ln (x) + c_{1}

Solution steps

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

Isolate

y^{'} (x) : y^{'} (x) = \frac{e ^{- x} - x - 1}{x}

If

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{e ^{- x} - x - 1}{x} d x

\int \frac{e ^{- x} - x - 1}{x} d x = Ei (- x) - x - ln (x) + c_{1}

y = Ei (- x) - x - ln (x) + c_{1}

Popular Examples

integral of sin(2x)-5cos(4x)

\int sin (2 x) - 5 cos (4 x) d x

integral of 10x^5e^{x^6}

\int 10 x^{5} e^{x^{6}} d x

(\partial)/(\partial x)(ln(e^x+e^{-x}))

\frac{\partial}{\partial x} (ln (e^{x} + e^{- x}))

(\partial)/(\partial x)(1/a e^{-(x/a)^2})

\frac{\partial}{\partial x} (\frac{1}{a} e^{- (\frac{x}{a})^{2}})

tangent of f(x)=8x^2+3x,\at x=-45 tangent of

f (x) = 8 x^{2} + 3 x, at x = - 45

Frequently Asked Questions (FAQ)

What is the solution for xy^'+(1+x)=e^{-x} ?
The solution for xy^'+(1+x)=e^{-x} is y=\Ei(-x)-x-ln(x)+c_{1}