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Popular >

y+4y^3=(y^4+2x)y^'

Solution

y + 4 y^{3} = (y^{4} + 2 x) y^{^{'}}

Solution

y = - \frac{- 8 x + c _{1} + ( 8 x - c _{1} ) ^{2} + 8 x}{2}, y = - - \frac{- 8 x + c _{1} + ( 8 x - c _{1} ) ^{2} + 8 x}{2}, y = - \frac{- 8 x + c _{1} - ( 8 x - c _{1} ) ^{2} + 8 x}{2}, y = - - \frac{- 8 x + c _{1} - ( 8 x - c _{1} ) ^{2} + 8 x}{2}

Solution steps

y + 4 y^{3} = (y^{4} + 2 x) y^{'} (x)

Solve using substitution:

y = - \frac{- 8 x + c _{1} + ( 8 x - c _{1} ) ^{2} + 8 x}{2}, y = - - \frac{- 8 x + c _{1} + ( 8 x - c _{1} ) ^{2} + 8 x}{2}, y = - \frac{- 8 x + c _{1} - ( 8 x - c _{1} ) ^{2} + 8 x}{2}, y = - - \frac{- 8 x + c _{1} - ( 8 x - c _{1} ) ^{2} + 8 x}{2}

y = - \frac{- 8 x + c _{1} + ( 8 x - c _{1} ) ^{2} + 8 x}{2}, y = - - \frac{- 8 x + c _{1} + ( 8 x - c _{1} ) ^{2} + 8 x}{2}, y = - \frac{- 8 x + c _{1} - ( 8 x - c _{1} ) ^{2} + 8 x}{2}, y = - - \frac{- 8 x + c _{1} - ( 8 x - c _{1} ) ^{2} + 8 x}{2}

Popular Examples

integral of x^2cos(kx)

\int x^{2} cos (k x) d x

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

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

integral of 2^{2x+1}

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

tangent of f(x)=2-3x^2,\at x=1 tangent of

f (x) = 2 - 3 x^{2}, at x = 1

derivative of (x^3^2)

\frac{d}{d x} ((x^{3})^{2})

Frequently Asked Questions (FAQ)

What is the solution for y+4y^3=(y^4+2x)y^' ?
The solution for y+4y^3=(y^4+2x)y^' is