What is the solution for -2y^{''}+2y^'=0,y(1)=-2,y^'(-2)=3 ?

The solution for -2y^{''}+2y^'=0,y(1)=-2,y^'(-2)=3 is y=-2-3e^3+3e^{2+t}

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-2y^{''}+2y^'=0,y(1)=-2,y^'(-2)=3

- 2 y^{^{''}} + 2 y^{^{'}} = 0, y (1) = - 2, y^{^{'}} (- 2) = 3

y = - 2 - 3 e^{3} + 3 e^{2 + t}

Solution steps

- 2 y^{''} (t) + 2 y^{'} (t) = 0

Solve linear ODE:

y = - 2 - 3 e^{3} + 3 e^{2 + t}

y = - 2 - 3 e^{3} + 3 e^{2 + t}

x \to 0 lim (\frac{x}{3 + x - 3})

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\frac{d}{d y} (y e^{x y} cos (2 x) - 2 e^{x y} sin (2 x) + 2 x)

What is the solution for -2y^{''}+2y^'=0,y(1)=-2,y^'(-2)=3 ?
The solution for -2y^{''}+2y^'=0,y(1)=-2,y^'(-2)=3 is y=-2-3e^3+3e^{2+t}