What is the solution for (tdQ)/(dt)+Q=t^4ln(t) ?

The solution for (tdQ)/(dt)+Q=t^4ln(t) is Q(t)=(t^4ln(t))/5-(t^4}{25}+\frac{c_{1})/t

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(tdQ)/(dt)+Q=t^4ln(t)

\frac{t d Q}{d t} + Q = t^{4} ln (t)

Q (t) = \frac{t ^{4} ln ( t )}{5} - \frac{t ^{4}}{25} + \frac{c _{1}}{t}

Solution steps

\frac{t d Q ( t )}{d t} + Q (t) = t^{4} ln (t)

Solve linear ODE:

Q (t) = \frac{t ^{4} ln ( t )}{5} - \frac{t ^{4}}{25} + \frac{c _{1}}{t}

Q (t) = \frac{t ^{4} ln ( t )}{5} - \frac{t ^{4}}{25} + \frac{c _{1}}{t}

\frac{d}{d x} (\frac{1}{e ^{x}})

f (x) = 2 x^{2} - 5 x + 1, at x = 2

\int \frac{x ^{3}}{4 - x ^{2}} d x

(- 3, 1), (2, - 1)

\frac{d}{d x} (lo g_{10} (x^{2}))

What is the solution for (tdQ)/(dt)+Q=t^4ln(t) ?
The solution for (tdQ)/(dt)+Q=t^4ln(t) is Q(t)=(t^4ln(t))/5-(t^4}{25}+\frac{c_{1})/t