What is the solution for (tdy)/(dt)+3y=t^4,t>0,y(1)=3 ?

The solution for (tdy)/(dt)+3y=t^4,t>0,y(1)=3 is y=(t^4)/7+(c_{1})/(t^3)

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(tdy)/(dt)+3y=t^4,t>0,y(1)=3

\frac{t d y}{d t} + 3 y = t^{4}, t > 0, y (1) = 3

y = \frac{t ^{4}}{7} + \frac{c _{1}}{t ^{3}}

Solution steps

\frac{t d y}{d t} + 3 y = t^{4}

Solve linear ODE:

y = \frac{t ^{4}}{7} + \frac{c _{1}}{t ^{3}}

y = \frac{t ^{4}}{7} + \frac{c _{1}}{t ^{3}}

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

\frac{d}{d x} (ln (1 - x^{3}))

\int 2 + \frac{1}{x ^{2}} d x

\frac{d}{d x} ((1 - 4 x)^{- 1})

g (x) = 6^{x}

What is the solution for (tdy)/(dt)+3y=t^4,t>0,y(1)=3 ?
The solution for (tdy)/(dt)+3y=t^4,t>0,y(1)=3 is y=(t^4)/7+(c_{1})/(t^3)