Ex 9.4, 4 - Chapter 9 Class 12 Differential Equations

Transcript

Ex 9.4, 4 show that the given differential equation is homogeneous and solve each of them. (𝑥^2−𝑦^2 )𝑑𝑥+2𝑥𝑦 𝑑𝑦=0 Step 1: Find 𝑑𝑦/𝑑𝑥 (𝑥^2−𝑦^2 )𝑑𝑥+2𝑥𝑦 𝑑𝑦=0 2xy dy = − (𝑥^2−𝑦^2 ) dx 2xy dy = (𝑦^2−𝑥^2 ) dx 𝒅𝒚/𝒅𝒙 = (𝒚^𝟐 − 𝒙^𝟐)/𝟐𝒙𝒚 Step 2: Putting F(x, y) = 𝑑𝑦/𝑑𝑥 and finding F(𝜆x, 𝜆y) F(x, y) = (𝑦^2 − 𝑥^2)/2𝑥𝑦 F(𝜆x, 𝜆y) = ((𝜆𝑦)^2−(𝜆𝑥)^2)/(2𝜆𝑥. 𝜆𝑦)= (𝑥^2 𝑦^2 −𝜆^2 𝑥^2)/(𝜆^2.2𝑥𝑦)= (𝜆^2 (𝑦^2 − 𝑥^2))/(𝜆^2.2𝑥𝑦) = (𝑦^2 −𝑥^2)/2𝑥𝑦 = F(x, y) ∴ F(𝜆x, 𝜆y) = F(x, y) = 𝜆° F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, 𝑑𝑦/𝑑𝑥 is a homogenous differential equation. Step 3: Solving 𝑑𝑦/𝑑𝑥 by putting y = vx Put y = vx. differentiating w.r.t.x 𝑑𝑦/𝑑𝑥 = x 𝑑𝑣/𝑑𝑥+𝑣𝑑𝑥/𝑑𝑥 𝒅𝒚/𝒅𝒙 = 𝒙 𝒅𝒗/𝒅𝒙 + v Putting value of 𝑑𝑦/𝑑𝑥 and y = vx in (1) 𝑑𝑦/𝑑𝑥 = (𝑦^2 − 𝑥^2)/2𝑥𝑦 (𝒙 𝒅𝒗)/𝒅𝒙+𝒗 = (〖(𝒙𝒗)〗^𝟐 − 𝒙^𝟐)/(𝟐𝒙(𝒙 𝒗)) (𝑥 𝑑𝑣)/𝑑𝑥+𝑣 = (𝑥^2 𝑣^2−𝑥^2)/(2𝑥^2 𝑣) (𝑥 𝑑𝑣)/𝑑𝑥 = (𝑥^2 𝑣^2 − 𝑥^2)/(2𝑥^2 𝑣) − v x 𝑑𝑣/𝑑𝑥 = (𝑥^2 𝑣^2 − 𝑥^2 − 2𝑥^2 𝑣^2)/(2𝑥^2 𝑣) x 𝑑𝑣/𝑑𝑥 = (〖−𝑥〗^2 𝑣^2 − 𝑥^2)/(2𝑥^2 𝑣) 𝑑𝑣/𝑑𝑥 = − 1/𝑥 ((𝑥^2 〖(𝑣〗^2 + 1))/(2𝑥^2 𝑣)) 𝑑𝑣/𝑑𝑥 = − 1/𝑥 ((𝑣^2 + 1)/2𝑣) (𝟐𝒗 𝒅𝒗)/(𝒗^𝟐 + 𝟏) = (−𝒅𝒙)/𝒙 Integrating both sides ∫1▒2𝑣/(𝑣^2 + 1) 𝑑𝑣 = ∫1▒(−𝑑𝑥)/𝑥 ∫1▒𝟐𝒗/(𝒗^𝟐 + 𝟏) 𝒅𝒗 = −𝒍𝒐𝒈⁡|𝒙|+𝑪 Putting t = v2 + 1 diff.w.r.t v. 𝑑/𝑑𝑣 (v2 + 1) = 𝑑𝑡/𝑑𝑣 2v = 𝑑𝑡/𝑑𝑣 dv = 𝒅𝒕/𝟐𝒗 Now, From (2) ∫1▒2𝑣/𝑡 𝑑𝑡/2𝑣 = − log|𝑥|+𝑐 (from (2)) ∫1▒𝒅𝒕/𝒕 = − log|𝒙|+𝒄 log|𝑡| = − log|𝑥|+𝑐 Putting t = v2 + 1 log|"v2 + 1" | = –log|"x" |+𝑐 log |"v2 + 1" | + log|"x" |=𝑐 log |"x(v2 + 1)" | = C Putting v = 𝑦/𝑥 log |[(𝑦/𝑥)^2+1]𝑥|=𝑐 log|(𝑦^2 + 𝑥^2)/𝑥^2 ×𝑥|=𝑐 Putting 𝑐 = log c log |(𝒚^𝟐 + 𝒙^𝟐)/𝒙| = log c 𝑦^2+𝑥^2 = cx 𝒙^𝟐+𝒚^𝟐=𝒄𝒙 is the general solution of the given differential equation.