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Popular >

ty^'+2y=1

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

(dy}{dx}=\frac{(x^3))/y

\frac{d y}{d x} = \frac{( x ^{3} )}{y}

(dx)/(dt)=-2tx^2

\frac{d x}{d t} = - 2 t x^{2}

(x\frac{dy)/(dx)-y}{xy}-1+7(dy)/(dx)=e^x

\frac{x \frac{d y}{d x} - y}{x y} - 1 + 7 \frac{d y}{d x} = e^{x}

y^{^{'}} = y t, y (0) = 2

e^{-2y}(dy)/(dx)=x-2,y(1)=0

e^{- 2 y} \frac{d y}{d x} = x - 2, y (1) = 0

Frequently Asked Questions (FAQ)

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