Ex 9.4, 7 - Chapter 9 Class 12 Differential Equations

Transcript

Ex 9.4, 7 Show that the given differential equation is homogeneous and solve each of them. {𝑥𝑐𝑜𝑠(𝑦/𝑥)+𝑦 sin⁡(𝑦/𝑥) }𝑦 𝑑𝑥={𝑦𝑠𝑖𝑛(𝑦/𝑥)−𝑥 cos⁡(𝑦/𝑥) }𝑥 𝑑𝑦 Step 1: Find 𝑑𝑦/𝑑𝑥 {𝑥𝑐𝑜𝑠(𝑦/𝑥)+𝑦 sin⁡(𝑦/𝑥) }𝑦 𝑑𝑥={𝑦𝑠𝑖𝑛(𝑦/𝑥)−𝑥 cos⁡(𝑦/𝑥) }𝑥 𝑑𝑦 𝒅𝒚/𝒅𝒙=((𝒙 𝒄𝒐𝒔⁡(𝒚/𝒙) + 𝒚 𝒔𝒊𝒏⁡(𝒚/𝒙))/(𝒚 𝒔𝒊𝒏⁡(𝒚/𝒙) − 𝒙 𝒄𝒐𝒔⁡〖 (𝒚/𝒙)〗 ))" " 𝒚/𝒙 Step 2: Put 𝑑𝑦/𝑑𝑥 = F (x, y) and find F(𝜆x, 𝜆y) F(x, y) = ((𝑥 cos⁡(𝑦/𝑥) + 𝑦 sin⁡(𝑦/𝑥))/(𝑦 sin⁡(𝑦/𝑥) − 𝑥 cos⁡〖 (𝑦/𝑥)〗 )) 𝑦/𝑥 F(𝜆x, 𝜆y) = ((𝜆𝑥 cos⁡( 𝜆𝑦/𝜆𝑥) + 𝜆𝑦 sin⁡( 𝜆𝑦/𝜆𝑥 ))/(𝜆𝑥 sin⁡( 𝜆𝑦/𝜆𝑥) − 𝜆𝑥 cos⁡〖 ( 𝜆𝑦/𝜆𝑥 )〗 )) 𝜆𝑦/𝜆𝑥 = ((𝑥 cos⁡(𝑦/𝑥) + 𝑦 sin⁡(𝑦/𝑥))/(𝑦 sin⁡(𝑦/𝑥) − 𝑥 cos⁡〖 (𝑦/𝑥)〗 ))" " 𝑦/𝑥 = 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 Putting 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=((𝑥 cos⁡(𝑦/𝑥) + 𝑦 sin⁡(𝑦/𝑥))/(𝑦 sin⁡(𝑦/𝑥) − 𝑥 cos⁡〖 (𝑦/𝑥)〗 ))" " 𝑦/𝑥 v + x 𝒅𝒗/𝒅𝒙 = (𝒙 𝒄𝒐𝒔⁡(𝒗𝒙/𝒙) + 𝒗𝒙 𝒔𝒊𝒏⁡(𝒗𝒙/𝒙))/(𝒗𝒙 𝒔𝒊𝒏⁡(𝒗𝒙/𝒙) − 𝒙 𝒄𝒐𝒔⁡(𝒗𝒙/𝒙) ) × 𝒗𝒙/𝒙 v + x 𝑑𝑣/𝑑𝑥 = [𝑥𝑐𝑜𝑠 𝑣 + 𝑣𝑥 𝑠𝑖𝑛 𝑣]𝑣/(𝑣𝑥 sin⁡〖𝑣 − 𝑥 cos⁡𝑣 〗 ) v + x 𝑑𝑣/𝑑𝑥 = 𝑥[𝑐𝑜𝑠 𝑣 + 𝑣 𝑠𝑖𝑛 𝑣]𝑣/𝑥[𝑣 𝑠𝑖𝑛 𝑣 −〖 cos〗⁡𝑣 ] v + x 𝑑𝑣/𝑑𝑥 = 𝑣[𝑐𝑜𝑠 𝑣 + 𝑣 𝑠𝑖𝑛 𝑣]/(𝑣 𝑠𝑖𝑛 𝑣 −〖 cos〗⁡𝑣 ) x ( 𝑑𝑣)/𝑑𝑥 = (𝑣 cos⁡𝑣 + 𝑣^2 sin⁡𝑣)/(𝑣 𝑠𝑖𝑛 𝑣 −〖 cos〗⁡𝑣 ) − v x ( 𝑑𝑣)/𝑑𝑥 = (𝑣 cos⁡𝑣 + 𝑣^2 sin⁡𝑣 − 𝑣(𝑣 sin⁡〖𝑣 〗− cos⁡𝑣 ))/(𝑣 𝑠𝑖𝑛 𝑣 −〖 cos〗⁡𝑣 ) x ( 𝑑𝑣)/𝑑𝑥 = (𝑣 cos⁡𝑣 + 𝑣^2 sin⁡𝑣 − 𝑣^2 sin⁡〖𝑣 + 𝑣 cos⁡𝑣 〗)/(𝑣 𝑠𝑖𝑛 𝑣 −〖 cos〗⁡𝑣 ) ( 𝑥𝑑𝑣)/𝑑𝑥 = (2𝑣 cos⁡𝑣)/(𝑣 𝑠𝑖𝑛 𝑣 −〖 cos〗⁡𝑣 ) ( 𝒅𝒗)/𝒅𝒙 = ( 𝟏)/𝒙 [(𝟐𝒗 𝒄𝒐𝒔⁡𝒗)/(𝒗 𝒔𝒊𝒏 𝒗 −〖 𝒄𝒐𝒔〗⁡𝒗 )] (𝑣 sin⁡〖𝑣 − cos⁡𝑣 〗)/(𝑣 〖 cos〗⁡𝑣 ) dv = 2 ( 𝑑𝑥)/𝑥 Integrating both sides. ∫1▒(𝒗 𝒔𝒊𝒏⁡〖𝒗 −𝒄𝒐𝒔⁡𝒗 〗)/(𝒗 〖 𝒄𝒐𝒔〗⁡𝒗 ) dv = 2 ( 𝒅𝒙)/𝒙 ∫1▒(𝑣 sin⁡〖𝑣 〗)/(𝑣 〖 cos〗⁡𝑣 ) dv − ∫1▒cos⁡𝑣/(𝑣 〖 cos〗⁡𝑣 ) dv = 2∫1▒𝑑𝑥/𝑥 ∫1▒sin⁡〖𝑣 〗/cos⁡𝑣 dv − ∫1▒𝑑𝑣/𝑣 dv = 2∫1▒𝑑𝑥/𝑥 ∫1▒tan⁡𝑣 dv − ∫1▒𝑑𝑣/𝑣 dv = 2∫1▒𝑑𝑥/𝑥 log⁡〖|sec⁡𝑥 |−log⁡|𝑣|=2 log⁡|𝑥| 〗 + C 𝒍𝒐𝒈⁡〖|𝒔𝒆𝒄⁡𝒙/𝒗|=𝒍𝒐𝒈⁡〖𝒙^𝟐 〗+𝒄 〗 log⁡〖|sec⁡𝑥/𝑣|=log⁡〖𝑥^2 〗+log⁡𝑐 〗 log⁡〖|sec⁡𝑥/𝑣|=log⁡〖𝑐𝑥^2 〗 〗 Put v = 𝑦/𝑥 log |𝐬𝐞𝐜⁡(𝒚/𝒙)/(𝒚/𝒙)| = log (cx2) |〖sec 〗⁡〖(𝑦/𝑥) 〗/(𝑦/𝑥)| = cx2 |〖sec 〗⁡〖(𝑦/𝑥) 〗 | = cx2 × 𝑦/𝑥 |〖𝒔𝒆𝒄 〗⁡(𝒚/𝒙) |= C 1/|cos⁡(𝑦/𝑥) | = C xy xy |𝑐𝑜𝑠(𝑦/𝑥)| = c1 xy 𝒄𝒐𝒔|𝒚/𝒙| = c1