Question

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Problem3. Solve the initial value problem y' = 5y + e -*y2 y(O)= 2

Answer 1

soli - Given y'= 5y + e 22 j2 - 0 460]=2 yl-5y = 21,527 Equation ☺ is Bernoullis Equation that is dy + py = Q yh I de Q = ē n =-2 Where PE-5 Now multiply Equation by y2 on both sides. чу! 59 (9) - см * * - у'- ду- -х yz dy _5y3 = 0 2 2 ② Put yo = v => differentiation 1. da ay² ds = dy an dae y en - 4 and 6 in Eqn② substitute Equation da § du - 50 = ēzx -> Now multiply By 3 on Eqh 6 Buduaar -3(50) = 3 ēa dv - 15u= 3 e2 da

O da Equation ® is linear differential Equation. that is du + pv=a. where = -15 a = 3 eat Integrating factor (I.F) = ( pda = 1 - 15 da VIF - ē5x Jduzu. General Solution NCI.F) = sq (I.F) dz+c vē 151 - 58ē22 (€152dute = 3 Jelle ante de=H. ..pean due . Je on 72 ante 2-15n = 3 -177 e - tc -17 ve 15n- -1122 1 E sccc 152 V = -3 -inn+ 152 '' + Ce F Tº= 22r + celu E y = [-sexn+ce15a] Substitute yeo=2 in Ean that is yx)=4 2=0, 4=2 2= 6-2 20 tlé®]" é = 1 2=(-2, +Cjxz - Take kube on both sides. 23-63, +C) = 8 +3 =( 1363 =(

'ľ in Egn ® substitute y = [-3ē + 139 ent] 17 yo = 11-3222 +1390152