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Popular >

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

Solution

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

Solution

y = \frac{e ^{2 + x + c_{1}}}{4 + 4 x + x ^{2}}

Solution steps

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

Solve separable ODE:

y = \frac{e ^{2 + x + c_{1}}}{4 + 4 x + x ^{2}}

y = \frac{e ^{2 + x + c_{1}}}{4 + 4 x + x ^{2}}

Popular Examples

laplacetransform e^s laplace transform

e^{s}

limit as x approaches c of 5

x \to c lim (5)

tangent of f(x)=x^4-5x^2+4,\at x=2 tangent of

f (x) = x^{4} - 5 x^{2} + 4, at x = 2

(dy)/(dx)-y/x =xe^x

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

integral of (x^2)/(x^2-x+1)

\int \frac{x ^{2}}{x ^{2} - x + 1} d x

Frequently Asked Questions (FAQ)

What is the solution for (2+x)y^'-xy=0 ?
The solution for (2+x)y^'-xy=0 is y=(e^{2+x+c_{1}})/(4+4x+x^2)