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Popular >

y^'-x(y-1)=0

Solution

y^{^{'}} - x (y - 1) = 0

Solution

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

Solution steps

y^{'} (x) - x (y - 1) = 0

Solve separable ODE:

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

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

Graph

Popular Examples

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

(dy)/(dx)=(9+17x)/(xy^2),x>0,y(1)=2

\frac{d y}{d x} = \frac{9 + 17 x}{x y ^{2}}, x > 0, y (1) = 2

solvefor t,e^{-4\H(t)}d\H(t)=e^{6t}dt solve for

t, e^{- 4 H (t)} d H (t) = e^{6 t} d t

(1+y^2)dx-y(x+x^3)dy=0

(1 + y^{2}) d x - y (x + x^{3}) d y = 0

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

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

Frequently Asked Questions (FAQ)

What is the solution for y^'-x(y-1)=0 ?
The solution for y^'-x(y-1)=0 is y=e^{(x^2)/2+c_{1}}+1