# Ex 9.4, 12 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Last updated at April 16, 2024 by Teachoo

Ex 9.4, 12 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : š„^2 šš¦+(š„š¦+š¦^2 ) šš„=0;š¦=1 When š„=1The differential equation can be written šs š„^2 šš¦ = ā(xy + y2) dx šš¦/šš„ = (ā(š„š¦ + š¦^2 ))/š„^2 "Let F(x, y) = " šš¦/šš„ " =" (ā(š„š¦ +š¦^2 ))/š„^2 Finding F(šx, šy) F(šx, šy) = (ā(šš„šš¦ + š^2 š¦^2 ))/ćš^2 š„ć^2 = (āš^2 (š„š¦ + š¦^2 ))/ćš^2 š„ć^2 = (ā(š„š¦ + š¦^2 ))/š„^2 = šĀ° F(x, y) = F(x, y) = (ā(š„š¦ +š¦^2 ))/š„^2 ā“ F(x, y) is a homogenous function of degree zero Putting y = vx Diff w.r.t. x š š/š š = x š š/š š + v Putting value of šš¦/šš„ and y = vx in (1) šš¦/šš„ = (ā(š„š¦ + š¦^2 ))/š„^2 v + (š š š)/š š = (ā(š(šš) + (šš)^š ))/š^š v + (š„ šš£)/šš„ = (ā(š„^2 š£ + š„^2 š£^2))/š„^2 v + (š„ šš£)/šš„ = āš„^2 ((š£ + š£^2))/š„^2 v + (š„ šš£)/šš„ = ā(v2 + v) (š„ šš£)/šš„ = ā v2 ā v ā v (š„ šš£)/šš„ = ā(š£^2+2š£) š š/(š^š + šš) = ā š š/š Integrating both sides ā«1āšš£/(š£^2 + 2š£) = āā«1āšš„/š„ ā«1āšš£/(š£^2 + 2š£) = ā log x + log c ā«1āšš£/(ć(š£ć^2 + 2š£ + 1) ā 1) = ā log x + log c ā«1āš š/((š + š)^š ā š^š ) = ā log x + log c š/š log (š + š ā š)/(š + š + š) = ā log x + log c 1/2 log š£/(š£ + 2) = ā log x + log c log āš£/ā(š£ + 2) = ā log x + log C log āš/ā(š + š) + log x = log C log (š„ āš£)/ā(š£ + 2) = log C Using ā«1āšš„/(š„^2 ā š^2 ) = 1/2š log |(š„ ā š)/(š„ + š)|+š¶ (š„āš£)/ā(š£ + 2) = C Putting value of v i.e š¦/š„ (šā(š/š))/ā(š/š + š) = C ā(š„^2 Ć š¦/š„)/ā(š¦/š„ + 2) = C āš„š¦/ā((š¦ + 2š„)/š„) = C (š„āš¦)/ā(š¦ + 2š„) = C š„āš¦ = Cā(š¦+2š„) Squaring both sides x2y = c2(y + 2x) Putting x = 1 & y = 1 in (2) 12(1) = C2(1 + 2) 1 = 3C2 C2 = š/š Putting value in (2) x2 y = 1/3(y + 2x) 3x2y = y + 2x y + 2x = 3x2y