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Popular >

(1+x)y^'-xy=0

Solution

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

Solution

y = \frac{e ^{1 + x + c_{1}}}{1 + x}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{1 + x + c_{1}}}{1 + x}

y = \frac{e ^{1 + x + c_{1}}}{1 + x}

Graph

Popular Examples

y^{^{'}} sec (x) = 5 y^{2}

(dy)/(dx)=3e^{x-y},y(0)=ln(9)

\frac{d y}{d x} = 3 e^{x - y}, y (0) = ln (9)

\frac{d v}{d t} = 1 - 2 v

(dy)/(dx)=((3e^x))/(3y+1)

\frac{d y}{d x} = \frac{( 3 e ^{x} )}{3 y + 1}

(dx)/(dt)=cos^2(x)

\frac{d x}{d t} = cos^{2} (x)

Frequently Asked Questions (FAQ)

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