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Popular >

y^'=-x(y+1),y(0)=2

Solution

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

Solution

y = 3 e^{- \frac{x ^{2}}{2}} - 1

Solution steps

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

Solve separable ODE:

y = 3 e^{- \frac{x ^{2}}{2}} - 1

y = 3 e^{- \frac{x ^{2}}{2}} - 1

Popular Examples

(\partial)/(\partial y)(2x+sqrt(y))

\frac{\partial}{\partial y} (2 x + y)

integral of 2x(4x^2-10)^2

\int 2 x (4 x^{2} - 10)^{2} d x

derivative of f(x)=-3x^2+5x+2 derivative of

f (x) = - 3 x^{2} + 5 x + 2

integral from 2 to 3 of x/((x^2-3)^2)

\int_{2}^{3} \frac{x}{( x ^{2} - 3 ) ^{2}} d x

integral of 4/((x^2-4))

\int \frac{4}{( x ^{2} - 4 )} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=-x(y+1),y(0)=2 ?
The solution for y^'=-x(y+1),y(0)=2 is y=3e^{-(x^2)/2}-1