Y X Ye Y-2x Dy Find the General Solution
Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at Sept. 14, 2018 by Teachoo
Find the particular solution of the differential equation :
ye y dx = (y 3 + 2xe y ) dy, y(0) = 1
OR
Show that (x − y)dy = (x + 2y)dx is a homogenous differential equation. Also, find the general solution of the given differential equation.
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Transcript
Question 19 Find the particular solution of the differential equation : yey dx = (y3 + 2xey) dy, y(0) = 1 Given equation yey dx = (y3 + 2xey) dy We try to put in form ππ¦/ππ₯ + Py = Q or ππ₯/ππ¦ + P1 x = Q1, yey dx = (y3 + 2xey) dy (π¦π^π¦)/((π¦3 + 2π₯π^π¦)) = ππ¦/ππ₯ ππ¦/ππ₯ = (π¦π^π¦)/((π¦3 + 2π₯π^π¦)) This is not of the form ππ¦/ππ₯ + Py = Q ∴ We find ππ₯/ππ¦ ππ₯/ππ¦ = ((π¦^3 + 2π₯π^π¦))/(π¦π^π¦ ) ππ₯/ππ¦ = π¦^3/(π¦π^π¦ )+(2π₯π^π¦)/(π¦π^π¦ ) ππ₯/ππ¦ = π¦^2/π^π¦ +2π₯/π¦ ππ₯/ππ¦−2π₯/π¦ = π¦^2/π^π¦ ππ₯/ππ¦+((−2)/π¦)π₯ = π¦^2/π^π¦ Comparing with ππ₯/ππ¦ + P1 x = Q1 Where P1 = (−2)/π¦ & Q1 = π¦^2/π^π¦ Find Integrating factor, IF = π^∫1▒〖π1 ππ¦〗 = π^∫1▒(−2ππ¦)/π¦ = π^(−2∫1▒ππ¦/π¦) = π^(−2 logπ¦ ) Using a log b = log ba = π^(〖logπ¦〗^(−2) ) = π¦^(−2) = 1/π¦^2 Solution is π₯ (IF) = ∫1▒〖(π1×πΌπΉ)ππ¦+π〗 π₯ × 1/π¦^2 = ∫1▒〖π¦^2/π^π¦ ×1/π¦^2 ππ¦+π〗 π₯/π¦^2 = ∫1▒〖1/π^π¦ ππ¦+π〗 π₯/π¦^2 = ∫1▒〖π^(−π¦) ππ¦+π〗 π₯/π¦^2 = −1 ×π^(−π¦)+π π₯/π¦^2 = 〖−π〗^(−π¦)+π π₯ = 〖−π¦^2 π〗^(−π¦)+ππ¦^2 We need to find particular solution, Putting x = 0, y = 1 0 = 〖−1^2 π〗^(−1)+π〖(1)〗^2 0 = 〖−π〗^(−1)+π π^(−1) =π c=π^(−1) c=1/π Putting value of c in (2) π₯ = 〖−π¦^2 π〗^(−π¦)+ππ¦^2 π₯ = 〖−π¦^2 π〗^(−π¦)+(1/π) π¦^2 π = 〖−π^π π〗^(−π)+π^π/π Question 19 Show that (x − y)dy = (x + 2y)dx is a homogenous differential equation. Also, find the general solution of the given differential equation. Theory To prove homogenous Step 1: Find ππ¦/ππ₯ Step 2: Put F(π₯ , π¦)=ππ¦/ππ₯ & Find F(ππ₯ ,ππ¦) Step 3: Then solve using by putting π¦=π£π₯ Finding ππ¦/ππ₯ (x − y)dy = (x + 2y)dx ππ¦/ππ₯=((π₯ + 2π¦)/(π₯ − π¦)) Now, Putting F(π₯ , π¦)=ππ¦/ππ₯ & Find F(ππ₯ ,ππ¦) Let F(π₯ , π¦)=((π₯ + 2π¦)/(π₯ − π¦)) Finding F(ππ ,ππ) F(ππ₯ ,ππ¦)=(ππ₯ + 2(ππ¦))/(ππ₯ −ππ¦) =π(π₯ + 2π¦)/(π (π₯ − π¦) ) =((π₯ + 2π¦))/(π₯ − π¦) = F(π₯ , π¦) Thus , F(ππ₯ ,ππ¦)="F" (π₯ , π¦)" " =π°" F" (π₯ , π¦)" " Thus , "F" (π₯ , π¦)" is Homogeneous function of degree zero" Therefore, the given Differential Equation is Homogeneous differential Equation Step 3: Solving ππ¦/ππ₯ by Putting π¦=π£π₯ ππ¦/ππ₯=((π₯ + 2π¦)/(π₯ − π¦)) Let π¦=π£π₯ So , ππ¦/ππ₯=π(π£π₯)/ππ₯ ππ¦/ππ₯=ππ£/ππ₯ . π₯+π£ ππ₯/ππ₯ ππ¦/ππ₯=ππ£/ππ₯ π₯+π£ Putting ππ¦/ππ₯ πnd π¦/π₯ ππ (π) π.π. ππ¦/ππ₯=(π₯ + 2π¦)/(π₯ − π¦) ππ£/ππ₯ π₯+π£= (π₯ + 2 π£π₯)/(π₯ − π£π₯) ππ£/ππ₯ π₯+π£= π₯(1 + 2π£)/π₯(1 − π£) ππ£/ππ₯ π₯+π£= (1 + 2π£)/(1− π£) ππ£/ππ₯ π₯= (1 + 2π£)/(1− π£)−π£ ππ£/ππ₯ . π₯= (1 + 2π£ − π£ (1− π£))/(1 −π£) ππ£/ππ₯ . π₯= (1 + 2π£ − π£ +〖 π£〗^2)/(1 − π£) ππ£/ππ₯ . π₯= (〖 π£〗^2 + π£ + 1)/(1 − π£) ππ£/ππ₯ π₯=−((〖 π£〗^2 + π£ + 1)/(π£ − 1)) ππ£((π£−1)/(π£^(2 )+ π£ + 1))=(− ππ₯)/π₯ Integrating Both Sides ∫1▒〖(π£ −1)/(π£^2 + π£ + 1) ππ£=∫1▒(−ππ₯)/π₯〗 ∫1▒〖(π£ −1)/(π£^2 + π£ + 1) ππ£=−∫1▒ππ₯/π₯〗 ∫1▒〖((π£ −1) ππ£)/(π£^2 + π£ + 1) ππ£〗=−log〖|π₯|〗 + π We can write π£^2+π£+1 = π£^2 + 1/2 . 2v + (1/2)^2+1−(1/2)^2 =(π£+1/2)^2 + 1 – 1/4 =(π£+1/2)^2+3/4 Putting π£^2+π£+1=(π£+1/2)^2+3/4 & π£−1=π£+π/π−π/π−1 =(π£+1/2)−3/2 ∫1▒((π£ + 1/2) − 3/2)/((π£ + 1/2)^2+ 3/4) ππ£=−log〖π₯+π〗 ∫1▒(π£ + 1/2)/((π£ + 1/2)^2+ 3/4) ππ£−3/2 ∫1▒1/((π£ + 1/2)^2+ 3/4) ππ£=−log〖π₯+π〗 Thus, I = I1 – (3 )/2 I2 Solving πΌ1 πΌ1=∫1▒((π£ + 1/2))/((π£ + 1/2)^2+ 3/4) ππ£ Put (π£+ 1/2)^2+ 3/4 =π‘ Diff. w.r.t. π£ π((π£ + 1/2)^2+ 3/4)/ππ£=ππ‘/ππ£ 2(π£+1/2)=ππ‘/ππ£ ππ£=ππ‘/2(π£ + 1/2) Putting value of v & dv in I1 πΌ1=∫1▒((π£ + 1/2))/π‘ ×ππ‘/2(π£ + 1/2) =1/2 ∫1▒ππ‘/π‘ =1/2 log |π‘| Putting π‘=(π£+ 1/2)^2+3/4 =1/2 πππ|(π£+ 1/2)^2+3/4| =1/2 πππ|π£^2+2π£ × 1/2 + 1/4 + 3/4| I1 =1/2 log〖 |π£^2+π£+1|〗 Solving π°π πΌ2=∫1▒ππ£/((π£ + 1/2)^2+3/4) =∫1▒ππ£/((π£ + 1/2)^2+(√3/2)^2 ) Put π‘=π£+1/2 Diff. w.r.t. π£ ππ‘/ππ£=1 ⇒ ππ‘=ππ£ = ∫1▒ππ‘/(π‘^2 + 〖 (√3/2)〗^2 ) =1/(√3/2) tan^(−1)〖π‘/(√3/2)〗 =2/√3 tan^(−1)〖2(π£ + 1/2)/√3〗 =2/√3 tan^(−1)((2π£ + 1)/√3) Hence I = πΌ1−3/2 πΌ2 I =1/2 log |π£^2+π£+1|−3/2 ×2/√3 tan^(−1)((2π£ +1)/√3) I =1/2 log |π£^2+π£+1|−√3 tan^(−1)((2π£ +1)/√3) Replacing v by (π¦ )/π₯ I =1/2 πππ|(π¦/π₯)^2+π¦/π₯+1|−√3 tan^(−1)((2π¦/π₯ + 1)/√3) I =1/2 πππ|π¦^2/π₯^2 +π¦/π₯+1|−√3 tan^(−1)((2π¦ + π₯)/(√3 π₯)) Putting Value of I in (2) 1/2 πππ|π¦^2/π₯^2 +π¦/π₯+1|−√3 tan^(−1)((2π¦ + π₯)/(√3 π₯))=−πππ|π₯|+π 1/2 πππ|π¦^2/π₯^2 +π¦/π₯+1|+πππ|π₯|=√3 tan^(−1)((2π¦ + π₯)/(√3 π₯))+π Multiplying Both Sides By 2 πππ|π¦^2/π₯^2 +π¦/π₯+1|+2 πππ|π₯|=2 √3 tan^(−1)((2π¦ + π₯)/(√3 π₯))+2π πππ|π¦^2/π₯^2 +π¦/π₯+1|+πππ|π₯|^2=2√3 tan^(−1)((2π¦ + π₯)/(√3 π₯))+2π Put 2π=π πππ[|π¦^2/π₯^2 +π¦/π₯+1| × |π₯^2 |]=2√3 tan^(−1)((2π¦ + π₯)/(√3 π₯))+π πππ|〖π₯^2 π¦〗^2/π₯^2 +(π₯^2 π¦)/π₯+π₯^2 |=2√3 tan^(−1)((π₯ + 2π¦)/(√3 π₯))+π πππ|π^π+ππ+π^π |=π√π 〖πππ〗^(−π)((π + ππ)/(√π π))+π Is the General Solution of the Differential Equation given
Y X Ye Y-2x Dy Find the General Solution
Source: https://www.teachoo.com/7210/2242/Question-19/category/CBSE-Sample-Paper-Class-12---2017-18/
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