What is the solution for y^'+y=e^{6t},y(0)=-1 ?

The solution for y^'+y=e^{6t},y(0)=-1 is y=(e^{6t})/7-8/(7e^t)

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y^'+y=e^{6t},y(0)=-1

y^{^{'}} + y = e^{6 t}, y (0) = - 1

y = \frac{e ^{6 t}}{7} - \frac{8}{7 e ^{t}}

Solution steps

y^{'} (t) + y = e^{6 t}

Solve linear ODE:

y = \frac{e ^{6 t}}{7} - \frac{8}{7 e ^{t}}

y = \frac{e ^{6 t}}{7} - \frac{8}{7 e ^{t}}

y^{^{'}} = sin (5 x)

h^{^{'}} = - h^{\frac{1}{3}}

\frac{d y}{d t} + e^{t} y = 23 e^{t}

\int \frac{ln ( x ^{25} )}{x} d x

\frac{\partial}{\partial y} (x^{5} + 2^{y} + x^{y})

What is the solution for y^'+y=e^{6t},y(0)=-1 ?
The solution for y^'+y=e^{6t},y(0)=-1 is y=(e^{6t})/7-8/(7e^t)