Question

Find the solution of the following initial value problem. y' (t) = 6tety(0) = 2, y'(0) = 4 y(t) = 1

Answer 1

Given y"(t) = 6 tet;9(0)=2,4'60)=4 Here y"(t) and y' (t) are derivatives of y(t) we can get y(t) from these derivatives using the Antidiarivaties (integrals) Sy"(t) dt y'CE) -> Sy'Ce)dt = y(t) di le. so, consider y Syn(t)dt y"(H) = 6tet Integrate both sides with t Sotet dt Iskem dn= ksfmda; Küconstanty » yl(t) = 6S tetat lets first solue Stet at C Scanned with CamScanner

Stet de let u= t derivate both sides ; Integrate both sides Suldt setat dt » üt uz l vzet Itlay ..or-solv. Tread an ean Now, Soula • Steldt - tet - sietat Beeldt - tet et et te y so, from 3 y(t) = 6 Stet de » y' (t) = 6[tet_et] tc (is integral constant CŚcanned with CamScanner

Given yl(o)=4 i... when too y'(4) = so, when tio co) y (0) - e) tc olo 6 o Co-DtC -6AC > vous, Calo y'(t) = otet-get to. Integrate both sides with t. sykt dt = Slotet - Get +10)dt » y(t) = 6Stethosethio Stadt from ③ we have Steedt= tet_et. * 40€) = ctet e-set tiot to [sendt- scanned with CamScanner EnA nail

is the required solution » y(t) = 6tet get betteotto d is integral constant »y(t) = otet_lzet tiot to Gillen y (o)=2 le. When t=of y(t) =2. so when tzo » 4(0)=G(o) e 12e +1000) td o al2010 totd d 2 4 12 (dziu :y(t) = stet izet tlot t14 Scanned with CamScanner