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Popular >

y^'=4x+xy

Solution

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

Solution

y = e^{\frac{x ^{2}}{2} + c_{1}} - 4

Solution steps

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

Solve separable ODE:

y = e^{\frac{x ^{2}}{2} + c_{1}} - 4

y = e^{\frac{x ^{2}}{2} + c_{1}} - 4

Graph

Popular Examples

y^'(x)=(15+3x)/(3y^2-12y),y(0)=2

y^{^{'}} (x) = \frac{15 + 3 x}{3 y ^{2} - 12 y}, y (0) = 2

\frac{d x}{2 x} = d t

(2dy)/y =(2cos(2x))/(sin(2x))dx

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

(dv)/(dt)=9.81-10v,v(0)=0

\frac{d v}{d t} = 9.81 - 10 v, v (0) = 0

y^3(dy)/(dx)=2x^4y^4-4x^4

y^{3} \frac{d y}{d x} = 2 x^{4} y^{4} - 4 x^{4}

Frequently Asked Questions (FAQ)

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