Question

1.  Find the particular solution of the differential equation

dydx+ycos(x)=2cos(x)dydx+ycos⁡(x)=2cos⁡(x)

satisfying the initial condition y(0)=4y(0)=4.

2. Solve the following initial value problem:

8dydt+y=32t8dydt+y=32t

with y(0)=6.y(0)=6.

(1 point) Find the particular solution of the differential equation dy + y cos(x) = 2 cos(z) satisfying the initial condition y(0) = 4. Answer: y= 2+2e^(-sin(x)) Your answer should be a function of x.

Answer 1

O find the posticulo sol. & doff.ca + y cas> = 2005 , 40)=4 29 + cosx = 2a9x dy +cosx)y=2cosbx, whéh is a linea difs.eq. He rex)cos, QcX)= 200sx IF Spanda Scosa da Sinc gen sdiff lineca dijet.co.ie y Spoondx = Sospeso dada+c y sin = (acosx.e Sinas docte ut sinx=t =) cosa doadt = Saeedt + c zette y.e Sinac = 2e sinac 9: e nz geSin - .. y: 2+ ce Sin given y Co=4 by => -Sino 4 - 2+c.e =) 4:2+CC Veº) 42+c = c=2 :y=/2+2e Sine

② solve the sup & dry=3et, with yco) = 6 sd 8dx+y = 325 dy + 3 4 = 4t (divide ou both sides with 8) which is a linear difs.eq. Hee Pct) = & Qct)= 4 . 9:F: SPCdt født t's t/8 tlo. gen. sol. of Linear diff. q. 18 4. Spct)dt = Sa.esActs at at + c gets - Shtete dette yetle = 4 Stettette – Sind Stette at the test - Sct. estis dat tls = tes - Cliette stet18 _ ssetedt +8tetle - se tre = 8tetl8_64etle from @yells - 4[8tetlen 64 418] + c yetle satel_256&t! + c y ets[Batetle 25624c] T8

given y(o)=6y =) 6 = 32 (0)- 256+c-e G = - 256+ CCI) =):C: 262 -t/s .: 9 - 32t - 256 + 262.e tle .: y = 32(t-8) +262e