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

The solution for y^'+2y=3x,y(0)=3 is y=(3(2x-1))/4+(15)/(4e^{2x)}

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

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

y = \frac{3 ( 2 x - 1 )}{4} + \frac{15}{4 e ^{2 x}}

Solution steps

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

Solve linear ODE:

y = \frac{3 ( 2 x - 1 )}{4} + \frac{15}{4 e ^{2 x}}

y = \frac{3 ( 2 x - 1 )}{4} + \frac{15}{4 e ^{2 x}}

\int 1 + sin (x) d x

x \to π lim (- 1)

\frac{\partial}{\partial x} (\frac{a ^{3} b ^{4}}{c ^{2} x ^{4}})

y^{^{''}} + 9 y = - sin (3 t), y (0) = 0, y^{^{'}} (0) = \frac{1}{6}

\int (x^{2} + 2 x) cos (5 x) d x

What is the solution for y^'+2y=3x,y(0)=3 ?
The solution for y^'+2y=3x,y(0)=3 is y=(3(2x-1))/4+(15)/(4e^{2x)}