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Popular >

y^'=8y^2+xy^2,y(0)=1

Solution

y^{^{'}} = 8 y^{2} + x y^{2}, y (0) = 1

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

4(sin(t)y^'+cos(t)y)=cos(t)sin^5(t)

4 (sin (t) y^{^{'}} + cos (t) y) = cos (t) sin^{5} (t)

tangent of (2x-1)/(x+3),\at x=4 tangent of

\frac{2 x - 1}{x + 3}, at x = 4

integral of (2x^3)/(36x^2+1)

\int \frac{2 x ^{3}}{36 x ^{2} + 1} d x

integral of (5x^2+x+13)/(x^3-2x^2+3x-6)

\int \frac{5 x ^{2} + x + 13}{x ^{3} - 2 x ^{2} + 3 x - 6} d x

integral of sin^3(θ)cos^6(θ)

\int sin^{3} (θ) cos^{6} (θ) d θ

Frequently Asked Questions (FAQ)

What is the solution for y^'=8y^2+xy^2,y(0)=1 ?
The solution for y^'=8y^2+xy^2,y(0)=1 is y=-2/(16x+x^2-2)