Question

(3 points) (a) Find the general solution to y′′+2y′=0. Give your answer as y=... . In your answer, use c1 and c2 to denote arbitrary constants and x the independent variable. Enter c1 as c1 and c2 as c2.

(3 points) (a) Find the general solution to y" + 2y' = 0. Give your answer as y=... . In your answer, use c1 and c2 to denote arbitrary constants and x the independent variable. Enter cı as c1 and C2 as c2. help (equations) (b) Find the particular solution that satisfies y(0) = 1 and y'(O) = 1. help (equations)

Answer 1

(a) To Solve &- y"+ay'=o - (1) 2 (n) Put n Y=l, y'=rqy" and you arh then corresponding characteristic equation becomes for U is pat arzo r(r+a)=0 =)| P=0,- 27 Roots are real and distinct therefor Solution of (1) is 6 x 31 960) = C, e" + Geen - Od (-2) y (H) =c, et qe -2x Enter this Y(") = Gt ca e 1 for First Blank Now (b) initial are conditions given - atcelu as y(0)=1 and y(0)=1 from (a) yox) = 6+ Cae 1 y'(x) = 6,+ Ca e .(-2) -) y'(x) = 6; -Qzē -Qu

Using initial value y (0)=1 -200) =) Cit Ca e = 1 a) 6+ C2 = 1 - Q2 (as e 21) By using y'(0) = 1 =) 6.-26g 22 (0) - 1 -) C1-221- (3) On Subtracting (2)-(3) we have (91+C2)-(4-269) = 1-1 1 362 = 0 = [ca=o] Using Cazo in (2) - 9,=1-C2 = Ta=17 Hence Particular Solution becomes . y(x) = 1 + Oo é an =) Ty(x) = 1 1 for Second Blank enter I becomes Иence

Rules for finding solution for busilary .. - Equation Cor complementary solution case-l's Roots are real and Dietinet 1 Y = cient la planta.tene dore E where linker ...en are arbitrary constant. E Case- 1 Roote are real and Repeated . y z oed (etcan text --tend) Case (1) Roote are B Complex and Distinet. Chozi otiß, dziui - idem - Yze [licooby + la Sin Bu] + (3 e stitchen case-fius Roote are complex and Repeated C = d+lß , (3,42 QZ iß, es ido. . ich y = eart [1614 (23) Corpu + ((37 Cum) sinou) tes erst anatenedoa Case- (V) Roote are of the form dzie y = Bax [co cosh fox +62 316 10 w07
last page represent rule for general solution according to different roots of characteristic equation.