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Popular >

y^'=(x+1)y

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

5y^4(dy)/(dx)=4x

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

y^'-x^3y^2=4x^3

y^{^{'}} - x^{3} y^{2} = 4 x^{3}

(dy)/(dt)=yt^2-1.1y

\frac{d y}{d t} = y t^{2} - 1.1 y

(dy)/(dx)=(2x)/(3y)e^{x^2-3y^2},y(1)=-1

\frac{d y}{d x} = \frac{2 x}{3 y} e^{x^{2} - 3 y^{2}}, y (1) = - 1

((x^2-9))/x (dy)/(dx)+y=0

\frac{( x ^{2} - 9 )}{x} \frac{d y}{d x} + y = 0

Frequently Asked Questions (FAQ)

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