Answer:
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}}}
