What is the solution for ty^'+7y=t^2-t+1,y(1)= 1/7 ?

The solution for ty^'+7y=t^2-t+1,y(1)= 1/7 is y=(t^2)/9-t/8+1/7+1/(72t^7)

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ty^'+7y=t^2-t+1,y(1)= 1/7

t y^{^{'}} + 7 y = t^{2} - t + 1, y (1) = \frac{1}{7}

y = \frac{t ^{2}}{9} - \frac{t}{8} + \frac{1}{7} + \frac{1}{72 t ^{7}}

Solution steps

t y^{'} (t) + 7 y = t^{2} - t + 1

Solve linear ODE:

y = \frac{t ^{2}}{9} - \frac{t}{8} + \frac{1}{7} + \frac{1}{72 t ^{7}}

y = \frac{t ^{2}}{9} - \frac{t}{8} + \frac{1}{7} + \frac{1}{72 t ^{7}}

\int tan^{2} (d) x d x

\frac{d}{d x} (ln (x))

x \to 2 - lim (- \frac{1}{x ^{2} - 4})

f (x) = 3 x - \frac{1}{x}

\frac{d}{d x} (0.04 e^{x} x^{2})

What is the solution for ty^'+7y=t^2-t+1,y(1)= 1/7 ?
The solution for ty^'+7y=t^2-t+1,y(1)= 1/7 is y=(t^2)/9-t/8+1/7+1/(72t^7)