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Popular >

x^4y^'=xy+2x

Solution

x^{4} y^{^{'}} = x y + 2 x

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

x^2y^'-cos(2y)=1

x^{2} y^{^{'}} - cos (2 y) = 1

(1+x^2)(dy)/(dx)+2xy=x

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

y^'-6y=0,y(-1)=10

y^{^{'}} - 6 y = 0, y (- 1) = 10

(dy)/(dx)=-(x+xy^2)/(y+x^2y)

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

(4x^2y^2-x^2+4xy^2-x)dx+(2xy+x+2y+1)dy=0

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

Frequently Asked Questions (FAQ)

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