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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

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

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

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

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

(dy)/(dx)=(x+2)/(2y)

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

y^'-2xy^3=0,y(2)=1

y^{^{'}} - 2 x y^{3} = 0, y (2) = 1

(dy)/(dt)=0.029y(1-y/5)

\frac{d y}{d t} = 0.029 y (1 - \frac{y}{5})

Frequently Asked Questions (FAQ)

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