Question

dy/dx=e^2x+(1+2e^x)y+y²,y1=-e^x​

Answer 1

Answer:

Step-by-step explanation:

Misc 9 Find the particular solution of the differential equation (1 + ^2) dy + (1 + ^2) ex dx = 0, given that y = 1 when x = 0. Given (1 + e2x) dy + (1 + y2)^ dx = 0 (1 + e2x) dy = −(1 + y2)^ dx / = (−(1 + ^2 ).^)/(1 + 2) /(1 + )^2 = (−^ )/(1 + 2) Integrating both sides ∫1▒/〖(1 + )〗^2 = ∫1▒( )/〖1 + 〗^2 …(1) Let t = ex Diff w.r.t.x /=^ /= ∴ Our equation becomes ∫1▒/〖1 + 〗^2 = −∫1▒〖( )/(1 + ^2 ) ( )/( )〗 ∫1▒/〖1 + 〗^2 = −∫1▒〖( )/(1 + ^2 ) 〗 tan^(−1)⁡=−tan^(−1)⁡+ Putting back value of t = ex 〖〗^(−)⁡=−〖〗^(−)⁡(^ )+ (As ∫1▒/〖1 + 〗^2 =tan^(−1)⁡) …(2) Given that y = 1 when x = 0 Put y = 1 and x = 0 in equation (2) tan^(−1)⁡〖(1)〗=−tan^(−1)⁡(^ )+ tan^(−1)⁡1=−tan^(−1)⁡+ tan^(−1)⁡1+tan^(−1)⁡1= 2 〖〗^(−)⁡= 2 × /= 2 × /2= C = /2. Putting value of C in (2) tan^(−1)⁡=−tan^(−1)⁡(^ )+ tan^(−1)⁡=−tan^(−1)⁡(^ )+" " /2 〖〗^(−)⁡+〖〗^(−)⁡(^ )=" " / is the required particular solution.