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Popular >

y^'=(xy^2)/8+y^2

Solution

y^{^{'}} = \frac{x y ^{2}}{8} + y^{2}

Solution

y = - \frac{16}{x ^{2} + 16 x + c _{1}}

Solution steps

y^{'} (x) = \frac{x y ^{2}}{8} + y^{2}

Solve separable ODE:

y = - \frac{16}{x ^{2} + 16 x + c _{1}}

y = - \frac{16}{x ^{2} + 16 x + c _{1}}

Graph

Popular Examples

(dy)/(dx)=x^2(y-6),y(0)=8

\frac{d y}{d x} = x^{2} (y - 6), y (0) = 8

(dy)/(dx)-2cos(x)y=0

\frac{d y}{d x} - 2 cos (x) y = 0

y^{1/2}y^'=1-y^{3/2}

y^{\frac{1}{2}} y^{^{'}} = 1 - y^{\frac{3}{2}}

(dy)/(dx)=(x^2y)

\frac{d y}{d x} = (x^{2} y)

(dy)/(dx)=((5x^6))/(y^2),y(0)=1

\frac{d y}{d x} = \frac{( 5 x ^{6} )}{y ^{2}}, y (0) = 1

Frequently Asked Questions (FAQ)

What is the solution for y^'=(xy^2)/8+y^2 ?
The solution for y^'=(xy^2)/8+y^2 is y=-(16)/(x^2+16x+c_{1)}