Ex 9.4

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

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 šš¦+(š„š¦+š¦^2 ) šš„=0;š¦=1 When š„=1The differential equation can be written šs š„^2 šš¦ = ā(xy + y2) dx šš¦/šš„ = (ā(š„š¦ + š¦^2 ))/š„^2 "Let F(x, y) = " šš¦/šš„ " =" (ā(š„š¦ +š¦^2 ))/š„^2 Finding F(šx, šy) F(šx, šy) = (ā(šš„šš¦ + š^2 š¦^2 ))/ćš^2 š„ć^2 = (āš^2 (š„š¦ + š¦^2 ))/ćš^2 š„ć^2 = (ā(š„š¦ + š¦^2 ))/š„^2 = šĀ° 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 šš¦/šš„ and y = vx in (1) šš¦/šš„ = (ā(š„š¦ + š¦^2 ))/š„^2 v + (š šš)/šš = (ā(š(šš) + (šš)^š ))/š^š v + (š„ šš£)/šš„ = (ā(š„^2 š£ + š„^2 š£^2))/š„^2 v + (š„ šš£)/šš„ = āš„^2 ((š£ + š£^2))/š„^2 v + (š„ šš£)/šš„ = ā(v2 + v) (š„ šš£)/šš„ = ā v2 ā v ā v (š„ šš£)/šš„ = ā(š£^2+2š£) šš/(š^š + šš) = ā šš/š Integrating both sides ā«1āšš£/(š£^2 + 2š£) = āā«1āšš„/š„ ā«1āšš£/(š£^2 + 2š£) = ā log x + log c ā«1āšš£/(ć(š£ć^2 + 2š£ + 1) ā 1) = ā log x + log c ā«1āš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 āš£/ā(š£ + 2) = ā log x + log C log āš/ā(š + š) + log x = log C log (š„ āš£)/ā(š£ + 2) = log C Using ā«1āšš„/(š„^2 ā š^2 ) = 1/2š log |(š„ ā š)/(š„ + š)|+š¶ (š„āš£)/ā(š£ + 2) = C Putting value of v i.e š¦/š„ (šā(š/š))/ā(š/š + š) = C ā(š„^2 Ć š¦/š„)/ā(š¦/š„ + 2) = C āš„š¦/ā((š¦ + 2š„)/š„) = C (š„āš¦)/ā(š¦ + 2š„) = C š„āš¦ = 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