Misc 4 - Chapter 9 Class 12 Differential Equations (Term 2)

Transcript

Misc 4 Prove that 𝑥^2−𝑦^2=𝑐(𝑥^2+𝑦^2 )^2 is the general solution of differential equation (𝑥^3−3𝑥𝑦^2 )𝑑𝑥=(𝑦^3−3𝑥^2 𝑦)𝑑𝑦, where 𝑐 is a parameter . Given differential equation (𝑥^3−3𝑥𝑦^2 )𝑑𝑥=(𝑦^3−3𝑥^2 𝑦)𝑑𝑦 (𝑥^3 − 3𝑥𝑦^2)/(𝑦^3 − 3𝑥^2 𝑦)=𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥=(𝑥^3 − 3𝑥𝑦^2)/(𝑦^(3 )− 3𝑥^2 𝑦) 𝑑𝑦/𝑑𝑥=(𝑥^3 (1 − (3𝑥𝑦^2)/𝑥^3 ))/(𝑦^(3 ) (1 −(3𝑥^2 𝑦)/𝑦^3 ) ) 𝑑𝑦/𝑑𝑥=(𝑥^3 (1 − (3𝑦^2)/𝑥^2 ))/(𝑦^(3 ) (1 −(3𝑥^2)/𝑦^2 ) ) 𝑑𝑦/𝑑𝑥=(𝑥/𝑦)^3×((1 − 3(𝑦/𝑥)^2 ))/((1 − 3(𝑥/𝑦)^2 ) ) Putting y = vx. Differentiating w.r.t. x 𝑑𝑦/𝑑𝑥 = 𝑥 𝑑𝑣/𝑑𝑥 + 𝑣 Putting value of 𝑑𝑦/𝑑𝑥 and y = vx in (1) 𝑥 𝑑𝑣/𝑑𝑥+𝑣 =(1/𝑣)^3×((1 − 3𝑣^2 ))/((1 − 3(1/𝑣)^2 ) ) …(1) 𝑥 𝑑𝑣/𝑑𝑥+𝑣 =1/𝑣^3 ×((1 − 3𝑣^2 ))/(((𝑣^2 − 3)/𝑣^2 ) ) 𝑥 𝑑𝑣/𝑑𝑥+𝑣 =1/𝑣×((1 − 3𝑣^2 ))/((𝑣^2 − 3) ) 𝑥 𝑑𝑣/𝑑𝑥=1/𝑣×((1 − 3𝑣^2 ))/((𝑣^2 − 3) )−𝑣 𝑥 𝑑𝑣/𝑑𝑥=1/𝑣×((1 − 3𝑣^2 ) − 𝑣 × 𝑣 (𝑣^2 − 3))/((𝑣^2 − 3) ) 𝑥 𝑑𝑣/𝑑𝑥=1/𝑣×(1 − 3𝑣^2 − 𝑣^4 + 3𝑣^2)/((𝑣^2 − 3) ) 𝑥 𝑑𝑣/𝑑𝑥=1/𝑣×(1 − 𝑣^4)/((𝑣^2 − 3) ) 𝑥 𝑑𝑣/𝑑𝑥=(1 − 𝑣^4)/((𝑣^3 − 3𝑣) ) (𝑣^3 −3𝑣)𝑑𝑣/((1 −𝑣^4 ) )=𝑑𝑥/𝑥 Integrating Both Sides ∫1▒〖(𝑣^3 −3𝑣 )/(1 − 𝑣^4 ) 𝑑𝑣〗=∫1▒𝑑𝑥/𝑥 ∫1▒〖(𝑣^3 −3𝑣 )/(1 − 𝑣^4 ) 𝑑𝑣〗=log⁡〖|𝑥|〗+𝐶 Let I = ∫1▒(𝑣^3 − 3𝑣)/(1 − 𝑣^4 ) 𝑑𝑣 Therefore, 𝐼 =log⁡〖|𝑥|+𝑐〗 …(1) Solving 𝑰 𝐼 =∫1▒〖(𝑣^3 −3𝑣 )/(1 − 𝑣^4 ) 𝑑𝑣〗 =∫1▒〖(𝑣^3 )/(1 − 𝑣^4 )−3∫1▒〖𝑣/(1 −〖 𝑣〗^4 ) 𝑑𝑣〗〗 𝐼 =∫1▒〖𝑣^3/(−𝑡) 𝑑𝑡/(4𝑣^3 ) −3∫1▒〖𝑣/(1 − 𝑝^2 ) 𝑑𝑝/2𝑣〗〗 Put 𝑣^4−1=𝑡 Diff. w.r.t. 𝑣 𝑑/𝑑𝑣 (𝑣^4−1)=𝑑𝑡/𝑑𝑣 4𝑣^3=𝑑𝑡/𝑑𝑣 𝑑𝑣=𝑑𝑡/(4𝑣^3 ) Put 𝑝=𝑣^2 Diff. w.r.t. 𝑣 𝑑𝑝/𝑑𝑣=2𝑣 𝑑𝑝/2𝑣=𝑑𝑣 …(2) 𝐼 =−1/4 ∫1▒〖 𝑑𝑡/𝑡−3/2 ∫1▒〖 𝑑𝑝/(1 − 𝑝^2 )〗〗 𝐼 =−1/4 ∫1▒〖 𝑑𝑡/𝑡+3/2 ∫1▒〖 𝑑𝑝/((𝑝^2 − 1^2 ) )〗〗 𝐼 = (−1)/( 4) log⁡𝑡+3/2 × 1/(2(1)) 𝑙𝑜𝑔((𝑝 − 1)/(𝑝 + 1)) Putting t = 𝑣^4 − 1 and p = v2 I = (−1)/4 log⁡〖(𝑣^4−1) 〗+3/4 log⁡〖((𝑣^2 − 1))/((𝑣^2 + 1))〗 I = 1/4 [−log⁡〖(𝑣^4−1)+3 𝑙𝑜𝑔 ((𝑣^2 − 1))/((𝑣^2 + 1))〗 ] I = 1/4 [−log⁡〖(𝑣^4−1)+ 𝑙𝑜𝑔 (𝑣^2 − 1)^3/(𝑣^2 + 1)^3 〗 ] I = 1/4 [𝑙𝑜𝑔 (𝑣^2 − 1)^3/(𝑣^2 + 1)^3 × 1/((𝑣^4 − 1))] (∫1▒〖𝑑𝑥/(𝑥^2 − 𝑎^2 )=1/2𝑎 𝑙𝑜𝑔((𝑥 − 𝑎)/(𝑥 + 𝑎)) 〗) I = 1/4 [log⁡〖1/((𝑣^4−1) )+ 𝑙𝑜𝑔 (𝑣^2 − 1)^3/(𝑣^2 + 1)^3 〗 ] I = 1/4 𝑙𝑜𝑔 1/((𝑣^4 − 1))×(𝑣^2 − 1)^3/(𝑣^2 + 1)^3 I = 1/4 𝑙𝑜𝑔 1/((𝑣^2 − 1)(𝑣^2 + 1))×(𝑣^2 − 1)^3/(𝑣^2 + 1)^3 I = 1/4 𝑙𝑜𝑔 (𝑣^2 − 1)^2/(𝑣^2 + 1)^4 I = 1/4 log⁡〖(((𝑣^2 − 1))/(𝑣^2 + 1)^2 )^2 〗 I = 1/4 × 2 log⁡〖((𝑣^2 − 1))/(𝑣^2 + 1)^2 〗 I = 1/2 log⁡〖((𝑣^2 − 1))/(𝑣^2 + 1)^2 〗 Putting back v = 𝑦/𝑥 I = 1/2 log⁡〖(((𝑦/𝑥)^2 − 1))/((𝑦/𝑥)^2 + 1)^2 〗 I = 1/2 log (((𝑦^2 − 𝑥2)/𝑥^2 )/((𝑦^2 + 𝑥2)/𝑥^2 )^2 ) I = 1/2 log [(𝑥2(𝑦^2 − 𝑥^2))/(𝑦^2 + 𝑥^2 )^2 ] Substituting value of I in (2) I = log |x| + c 1/2 log ⌈(𝑥2(𝑦2 − 𝑥2))/((𝑦2 + 𝑥2))⌉ = log |x| + c log ⌈(𝑥2(𝑦2 − 𝑥2))/((𝑥2 + 𝑦2))⌉ = 2 log |x| + 2c log ⌈(𝑥2(𝑦2 − 𝑥2))/((𝑥2 + 𝑦2))⌉ = log |x|2 + log c1 log ⌈(𝑥2(𝑦2 − 𝑥2))/((𝑥2 + 𝑦2))⌉ = log |x|2 + log c1 log ⌈(𝑥2(𝑦2 − 𝑥2))/(𝑥2 + 𝑦2)^2 ⌉ = log c1|x|2 Cancelling log (𝑥2(𝑦2 − 𝑥2))/(𝑥2 + 𝑦2)^2 = c1 x2 x2 (y2 − x2) = c1 x2 (x2 + y2)2 Cancelling x2 from both sides y2 − x2 = c1 (x2 + y2)2 x2 − y2 = c2 (x2 + y2)2 Hence proved