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Popular >

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

Solution

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

Solution

y = ln (e^{x} + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (e^{x} + c_{1})

y = ln (e^{x} + c_{1})

Popular Examples

(\partial)/(\partial x)(sin(x-y)+(x+y))

\frac{\partial}{\partial x} (sin (x - y) + (x + y))

(\partial)/(\partial x)(sqrt(4x+2y))

\frac{\partial}{\partial x} (4 x + 2 y)

tangent of f(x)=5tan(x),(pi/4 ,5)tangent of

f (x) = 5 tan (x), (\frac{π}{4}, 5)

derivative of sec(1+x^2)

\frac{d}{d x} (sec (1 + x^{2}))

integral of (63)/(1-cos(7x))

\int \frac{63}{1 - cos ( 7 x )} d x

Frequently Asked Questions (FAQ)

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