What is the solution for y^'+4y=3,2xy(0)=2 ?

The solution for y^'+4y=3,2xy(0)=2 is y=-(e^{-4x-c_{1}})/4+3/4

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y^'+4y=3,2xy(0)=2

y^{^{'}} + 4 y = 3, 2 x y (0) = 2

y = - \frac{e ^{- 4 x - c_{1}}}{4} + \frac{3}{4}

Solution steps

y^{'} (x) + 4 y = 3

Solve separable ODE:

y = - \frac{e ^{- 4 x - c_{1}}}{4} + \frac{3}{4}

y = - \frac{e ^{- 4 x - c_{1}}}{4} + \frac{3}{4}

(1 + 2 x) \frac{d y}{d x} = \frac{4 x ^{3}}{y}

2 x y y^{^{'}} - y^{2} = 1

y^{^{'}} = (x + 1) y^{2}, y (0) = \frac{1}{2}

u^{^{'}} = 5 e^{9 x - H (x)}

x (cos (y))^{2} d x + tan (y) d y = 0

What is the solution for y^'+4y=3,2xy(0)=2 ?
The solution for y^'+4y=3,2xy(0)=2 is y=-(e^{-4x-c_{1}})/4+3/4