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Popular >

y^'+rxy=0

Solution

y^{^{'}} + r x y = 0

Solution

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

Solution steps

y^{'} (x) + r x y = 0

Solve separable ODE:

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

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

Popular Examples

(((x+1)^2+6)/(4(x+1)))((dy)/(dx))=tan(y)

(\frac{( x + 1 ) ^{2} + 6}{4 ( x + 1 )}) (\frac{d y}{d x}) = tan (y)

(dy}{dx}=-\frac{ay)/x

\frac{d y}{d x} = - \frac{a y}{x}

y*y^'-2x=sin(x)

y \cdot y^{^{'}} - 2 x = sin (x)

(dy)/(dx)=0.125(y-3),y(0)=5

\frac{d y}{d x} = 0.125 (y - 3), y (0) = 5

e^x(dy)/(dx)-sec(y)-xsec(y)=0

e^{x} \frac{d y}{d x} - sec (y) - x sec (y) = 0

Frequently Asked Questions (FAQ)

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