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Popular >

y^'+ln(x)y=0

Solution

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

Solution

y = e^{x + c_{1}} x^{- x}

Solution steps

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

Solve separable ODE:

y = e^{x + c_{1}} x^{- x}

y = e^{x + c_{1}} x^{- x}

Popular Examples

(dy)/(dx)-e^{3x+2y}=0

\frac{d y}{d x} - e^{3 x + 2 y} = 0

yy^'+6x=0,y(0)=4

y y^{^{'}} + 6 x = 0, y (0) = 4

sqrt(x)(dy)/(dx)=e^{9y+sqrt(x)}

x \frac{d y}{d x} = e^{9 y + x}

(dy}{dt}=\frac{-t)/y

\frac{d y}{d t} = \frac{- t}{y}

y^{^{'}} = e^{x} sin (y)

Frequently Asked Questions (FAQ)

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