What is the solution for y^'+y=e^{9t},y(0)=-1 ?

The solution for y^'+y=e^{9t},y(0)=-1 is y=(e^{9t})/(10)-(11)/(10e^t)

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y^'+y=e^{9t},y(0)=-1

y^{^{'}} + y = e^{9 t}, y (0) = - 1

y = \frac{e ^{9 t}}{10} - \frac{11}{10 e ^{t}}

Solution steps

y^{'} (t) + y = e^{9 t}

Solve linear ODE:

y = \frac{e ^{9 t}}{10} - \frac{11}{10 e ^{t}}

y = \frac{e ^{9 t}}{10} - \frac{11}{10 e ^{t}}

x \to 2 lim (\frac{x + 3}{x + 2})

\frac{\partial}{\partial y} (x y^{3} + 4 y e^{x} - 2 y^{2})

\frac{\partial}{\partial y} (ln ((\frac{y}{x})^{2}))

f (x) = - 3 x^{3} - 2 x^{2} - x + 3, at x = - 1

\int \frac{x}{( 1 - x ^{2} ) ^{2}} d x

What is the solution for y^'+y=e^{9t},y(0)=-1 ?
The solution for y^'+y=e^{9t},y(0)=-1 is y=(e^{9t})/(10)-(11)/(10e^t)