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Popular >

xy^'=y+1+xy+x

Solution

x y^{^{'}} = y + 1 + x y + x

Solution

y = e^{x + c_{1}} x - 1

Solution steps

x y^{'} (x) = y + 1 + x y + x

Solve separable ODE:

y = e^{x + c_{1}} x - 1

y = e^{x + c_{1}} x - 1

Popular Examples

y^'=(sin(2x))/(cos(3y)),y(0)=1

y^{^{'}} = \frac{sin ( 2 x )}{cos ( 3 y )}, y (0) = 1

y^'(t)=(3+y)(1-y),y(0)=4

y^{^{'}} (t) = (3 + y) (1 - y), y (0) = 4

2yt(dy)+y^2(dt)=0

2 y t (d y) + y^{2} (d t) = 0

y^'+y=y(te^{t^2}+1),y(0)=1

y^{^{'}} + y = y (t e^{t^{2}} + 1), y (0) = 1

y^'=(y^2)/(x^3),y(2)=4

y^{^{'}} = \frac{y ^{2}}{x ^{3}}, y (2) = 4

Frequently Asked Questions (FAQ)

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