What is the solution for y^'+4y=1.4,y(0)=ce^{-4x}+0.35,y(0)=2 ?

The solution for y^'+4y=1.4,y(0)=ce^{-4x}+0.35,y(0)=2 is y=-0.25e^{-4x-c_{1}}+0.35

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y^'+4y=1.4,y(0)=ce^{-4x}+0.35,y(0)=2

y^{^{'}} + 4 y = 1.4, y (0) = c e^{- 4 x} + 0.35, y (0) = 2

y = - 0.25 e^{- 4 x - c_{1}} + 0.35

Solution steps

y^{'} (x) + 4 y = 1.4

Solve separable ODE:

y = - 0.25 e^{- 4 x - c_{1}} + 0.35

y = - 0.25 e^{- 4 x - c_{1}} + 0.35

\frac{d y}{d x} = - \frac{5}{5} x^{7} y^{6}

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

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

y + x \cdot y^{^{'}} = 0

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

What is the solution for y^'+4y=1.4,y(0)=ce^{-4x}+0.35,y(0)=2 ?
The solution for y^'+4y=1.4,y(0)=ce^{-4x}+0.35,y(0)=2 is y=-0.25e^{-4x-c_{1}}+0.35