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Popular >

y^'+x^3y=0

Solution

y^{^{'}} + x^{3} y = 0

Solution

y = e^{- \frac{x ^{4}}{4} + c_{1}}

Solution steps

y^{'} (x) + x^{3} y = 0

Solve separable ODE:

y = e^{- \frac{x ^{4}}{4} + c_{1}}

y = e^{- \frac{x ^{4}}{4} + c_{1}}

Popular Examples

(dx)/(dy)=(6x5-2x+1)/(cos(y)+ey)

\frac{d x}{d y} = \frac{6 x 5 - 2 x + 1}{cos ( y ) + ey}

(dy)/(dt)=sqrt(y)

\frac{d y}{d t} = y

(dh)/(dt)=-3/100*sqrt(h-18)

\frac{d h}{d t} = - \frac{3}{100} \cdot h - 18

y^{^{'}} - 4 x y^{2} = 0

(dy)/(dx)-e^{-y}sec^2(x)=0

\frac{d y}{d x} - e^{- y} sec^{2} (x) = 0

Frequently Asked Questions (FAQ)

What is the solution for y^'+x^3y=0 ?
The solution for y^'+x^3y=0 is y=e^{-(x^4)/4+c_{1}}