What is the solution for 2y^'+9y=3,y(0)=1 ?

The solution for 2y^'+9y=3,y(0)=1 is y= 1/3+2/(3e^{\frac{9t){2}}}

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2y^'+9y=3,y(0)=1

2 y^{^{'}} + 9 y = 3, y (0) = 1

y = \frac{1}{3} + \frac{2}{3 e ^{\frac{9 t}{2}}}

Solution steps

2 y^{'} (t) + 9 y = 3

Solve linear ODE:

y = \frac{1}{3} + \frac{2}{3 e ^{\frac{9 t}{2}}}

y = \frac{1}{3} + \frac{2}{3 e ^{\frac{9 t}{2}}}

x \to 0 - lim (- \frac{1}{x ^{2}})

\frac{d}{d x} ((x + 6) 2)

\int_{0}^{3} \frac{1}{x ^{2}} d x

f (x) = ln (x + 1)

x \to \infty lim (1 - \frac{2}{x})

What is the solution for 2y^'+9y=3,y(0)=1 ?
The solution for 2y^'+9y=3,y(0)=1 is y= 1/3+2/(3e^{\frac{9t){2}}}