Find the general solution to y' + 19t1$ y = t19 Use the variable I=∫et17dt where it occurs in your answer, since this integral is not easily computable. Note that the arbitrary constant C would come from actually computing the integral I, so you do not need to write it.

y=/frac{ - / frac{/sqrt[9]{18} Г (10/9, - / frac{19t^{18}}{18}) }{19^{10/9} {e^{/frac{19t^{18}}{18}}}

Step-by-step explanation:

From exercise we have:

C=0

y'+19t^{17} y=t^{19}

We know that a linear differential equation is written in the standard form:

y' + a (t) y = f (t)

we get that: a (t) = 19t^{17} and f (t) = t^{19}.

We know that the integrating factor is defined by the formula:

u (t) = e^{∫ a (t) dt}

⇒ u (t) = e^{∫ 19t^{17} dt} = e^{/frac{19t^{18}}{18}}

The general solution of the differential equation is in the form:

y=/frac{ ∫ u (t) f (t) dt + C}{u (t) }

y=/frac{ ∫ e^{/frac{19t^{18}}{18}} · t^{19} dt + 0}{e^{/frac{19t^{18}}{18}}}

y=/frac{ - / frac{/sqrt[9]{18} Г (10/9, - / frac{19t^{18}}{18}) }{19^{10/9} {e^{/frac{19t^{18}}{18}}}