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Popular >

(x^2+1)y^'+4xy=0

Solution

(x^{2} + 1) y^{^{'}} + 4 x y = 0

Solution

y = \frac{c _{1}}{x ^{4} + 2 x ^{2} + 1}

Solution steps

(x^{2} + 1) y^{'} (x) + 4 x y = 0

Solve separable ODE:

y = \frac{c _{1}}{x ^{4} + 2 x ^{2} + 1}

y = \frac{c _{1}}{x ^{4} + 2 x ^{2} + 1}

Graph

Popular Examples

(dy}{dx}=\frac{4y)/x ,y=2,x=1

\frac{d y}{d x} = \frac{4 y}{x}, y = 2, x = 1

y^2sec(3x)(dy)/(dx)=1,y(0)=0

y^{2} sec (3 x) \frac{d y}{d x} = 1, y (0) = 0

y^{^{'}} = sin (3 x) y^{2}

y^{^{'}} = y - (y)^{2}

(dy)/(dx)=(x-6)e^{-2y},y(6)=ln(6)

\frac{d y}{d x} = (x - 6) e^{- 2 y}, y (6) = ln (6)

Frequently Asked Questions (FAQ)

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