What is the solution for y^'=6y+2t-4,y(0)=3 ?

The solution for y^'=6y+2t-4,y(0)=3 is y=-t/3+11/18+43/18 e^{6t}

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y^'=6y+2t-4,y(0)=3

y^{^{'}} = 6 y + 2 t - 4, y (0) = 3

y = - \frac{t}{3} + \frac{11}{18} + \frac{43}{18} e^{6 t}

Solution steps

y^{'} (t) = 6 y + 2 t - 4

Solve linear ODE:

y = - \frac{t}{3} + \frac{11}{18} + \frac{43}{18} e^{6 t}

y = - \frac{t}{3} + \frac{11}{18} + \frac{43}{18} e^{6 t}

\int_{0}^{1} 1 - x d x

\frac{d}{d x} ((2 - x)^{\frac{3}{2}})

\frac{\partial}{\partial x} (4 x - 8 y^{3})

y^{^{''}} - 6 y = 0

\frac{x ^{2}}{x - 4}

What is the solution for y^'=6y+2t-4,y(0)=3 ?
The solution for y^'=6y+2t-4,y(0)=3 is y=-t/3+11/18+43/18 e^{6t}