Question

(1 point) Solve the separable initial value problem. 1. y' = 22 cos(x²)(1+y?), y(0) = 3 + y= 2. y' = ln(x)(1 + y²), y(1) = 3 → y

Answer 1

(D) y' 15 solution: 1. Given Initial Value problem is 2x cost?) (1 + y) ylol - 3 can write (۱) dy 2x cas (x²) (1+ y²) da separating the variables, we get oly 2x cos(x) dx 1 + y Integrating both sides, we get itys dy 5 2x cos(x) dx' +'c '< (2) where c is constant of integration ma - Now I dy tany ditan'w) = 1 1+y² dx 1 +2? (3) To evaluate 2x cas (x²) da 2x I U says above x² = t من integral substitute so that d (2²) du) dt dt 2a da 2 ada =) dt I- - Sint fax In cas (x²) dx s cosex?) (2x dx) s cast at ci o [sinx da **] jax cos(x²) dx = sunt a sint & Sin (012) sin (01²) - 4 (2+2 - con

y = Now given that 2-0 we is 15 we get tan 3 = - Ot (= tan 3 C substituting the values from (3) and (4) is (2), we get tant Sin (x²) te (5) it is ylo) = 3 i e at have y = 3 using this initial condition Sis 10²) TC sin lojte tan' 3 = ci sin o=o] in substituting the value of (5), we get tanly = Sin (x²) 3 y tan ( sin 12²) + tanta) is the required required solution of given Initial value problem. 12) Given Initial value problem is y' = enix (1+ y2) U yu) - 3 can write (1) dy en (x) 11+ g2) du variables we get + tant we ал separating the I dy encal da ity" Integrating both sides, get Vityr dy = 5 enco) dx ſ en cas dx + c where constant of integration (2) e is

Now (3) da To I ( say) we have I - - adx . I (4) we sity dy = tans ry) (.dltanta) ㅗ to evaluate Sen cas de Senixida 51. en (w) da Integrating by parts, we get I - en ca). /1 dr Sorte len (x) (1 dm) da en(x) -5. x entro | 1 da xenta a sen (dx = den (x3 - x substituting the values from (3) and (4) in (2), get tan x enero Given that yees ie we have y = 3; using this initial condition is (5), we get tan 3= 1 ln (1) - 1 + c tan' 3 110) - ) + c cenl1)=0) tan 3+1 O + c ca 1+tan 3 substituting the value of c in (5), we get tan y = x enca x+1+tantz ya tan ex en 10-x+ It tan '3) is the required solution of given Initial value problem. ) y (5) 2 + C = 3 aut 2-) > n >

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