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Popular >

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

Solution

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

Solution

y = - x - 1

Solution steps

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

Solve linear ODE:

y = - x - 1

y = - x - 1

Graph

Popular Examples

derivative of r/(sqrt(r^2+1))derivative of

\frac{r}{r ^{2} + 1}

integral of (-0.5x+450)

\int (- 0.5 x + 450) d x

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

limit as x approaches 0 of (x)^{tan(x)}

x \to 0 lim ((x)^{t a n (x)})

(\partial)/(\partial x)(y/(x^2+y^2+z^2))

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

Frequently Asked Questions (FAQ)

What is the solution for y^'=y+x,y(0)=-1 ?
The solution for y^'=y+x,y(0)=-1 is y=-x-1