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Popular >

ty^'+2y= 1/2

Solution

t y^{^{'}} + 2 y = \frac{1}{2}

Solution

y = - \frac{c _{1}}{4 t ^{2}} + \frac{1}{4}

Solution steps

t y^{'} (t) + 2 y = \frac{1}{2}

Solve separable ODE:

y = - \frac{c _{1}}{4 t ^{2}} + \frac{1}{4}

y = - \frac{c _{1}}{4 t ^{2}} + \frac{1}{4}

Graph

Popular Examples

y^'=(1-5x)y^2,y(0)=-1/8

y^{^{'}} = (1 - 5 x) y^{2}, y (0) = - \frac{1}{8}

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

\frac{d y}{d x} = e^{2 x + y}, y (0) = 0

x^2y^'=y^2e^{1/x}

x^{2} y^{^{'}} = y^{2} e^{\frac{1}{x}}

(dy)/(dx)=y+5,y(3)=2

\frac{d y}{d x} = y + 5, y (3) = 2

(dp)/(dt)=0.25p(1-p*10^{-4})

\frac{d p}{d t} = 0.25 p (1 - p \cdot 1 0^{- 4})

Frequently Asked Questions (FAQ)

What is the solution for ty^'+2y= 1/2 ?
The solution for ty^'+2y= 1/2 is y=-(c_{1})/(4t^2)+1/4