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 May 29, 2023 by Teachoo

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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 𝑑𝑦/𝑑𝑥 + Py = Q or 𝑑𝑥/𝑑𝑦 + P1 x = Q1,
yey dx = (y3 + 2xey) dy
(𝑦𝑒^𝑦)/((𝑦3 + 2𝑥𝑒^𝑦)) = 𝑑𝑦/𝑑𝑥
𝑑𝑦/𝑑𝑥 = (𝑦𝑒^𝑦)/((𝑦3 + 2𝑥𝑒^𝑦))
This is not of the form 𝑑𝑦/𝑑𝑥 + Py = Q
∴ We find 𝑑𝑥/𝑑𝑦
𝑑𝑥/𝑑𝑦 = ((𝑦^3 + 2𝑥𝑒^𝑦))/(𝑦𝑒^𝑦 )
𝑑𝑥/𝑑𝑦 = 𝑦^3/(𝑦𝑒^𝑦 )+(2𝑥𝑒^𝑦)/(𝑦𝑒^𝑦 )
𝑑𝑥/𝑑𝑦 = 𝑦^2/𝑒^𝑦 +2𝑥/𝑦
𝑑𝑥/𝑑𝑦−2𝑥/𝑦 = 𝑦^2/𝑒^𝑦
𝑑𝑥/𝑑𝑦+((−2)/𝑦)𝑥 = 𝑦^2/𝑒^𝑦
Comparing with 𝑑𝑥/𝑑𝑦 + P1 x = Q1
Where P1 = (−2)/𝑦 & Q1 = 𝑦^2/𝑒^𝑦
Find Integrating factor,
IF = 𝑒^∫1▒〖𝑃1 𝑑𝑦〗
= 𝑒^∫1▒(−2𝑑𝑦)/𝑦
= 𝑒^(−2∫1▒𝑑𝑦/𝑦)
= 𝑒^(−2 log𝑦 )
Using a log b = log ba
= 𝑒^(〖log𝑦〗^(−2) )
= 𝑦^(−2)
= 1/𝑦^2
Solution is
𝑥 (IF) = ∫1▒〖(𝑄1×𝐼𝐹)𝑑𝑦+𝑐〗
𝑥 × 1/𝑦^2 = ∫1▒〖𝑦^2/𝑒^𝑦 ×1/𝑦^2 𝑑𝑦+𝑐〗
𝑥/𝑦^2 = ∫1▒〖1/𝑒^𝑦 𝑑𝑦+𝑐〗
𝑥/𝑦^2 = ∫1▒〖𝑒^(−𝑦) 𝑑𝑦+𝑐〗
𝑥/𝑦^2 = −1 ×𝑒^(−𝑦)+𝑐
𝑥/𝑦^2 = 〖−𝑒〗^(−𝑦)+𝑐
𝑥 = 〖−𝑦^2 𝑒〗^(−𝑦)+𝑐𝑦^2
We need to find particular solution,
Putting x = 0, y = 1
0 = 〖−1^2 𝑒〗^(−1)+𝑐〖(1)〗^2
0 = 〖−𝑒〗^(−1)+𝑐
𝑒^(−1) =𝑐
c=𝑒^(−1)
c=1/𝑒
Putting value of c in (2)
𝑥 = 〖−𝑦^2 𝑒〗^(−𝑦)+𝑐𝑦^2
𝑥 = 〖−𝑦^2 𝑒〗^(−𝑦)+(1/𝑒) 𝑦^2
𝒙 = 〖−𝒚^𝟐 𝒆〗^(−𝒚)+𝒚^𝟐/𝒆
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 𝑑𝑦/𝑑𝑥
Step 2: Put F(𝑥 , 𝑦)=𝑑𝑦/𝑑𝑥 & Find F(𝜆𝑥 ,𝜆𝑦)
Step 3: Then solve using by putting 𝑦=𝑣𝑥
Finding 𝑑𝑦/𝑑𝑥
(x − y)dy = (x + 2y)dx
𝑑𝑦/𝑑𝑥=((𝑥 + 2𝑦)/(𝑥 − 𝑦))
Now, Putting F(𝑥 , 𝑦)=𝑑𝑦/𝑑𝑥 & Find F(𝜆𝑥 ,𝜆𝑦)
Let F(𝑥 , 𝑦)=((𝑥 + 2𝑦)/(𝑥 − 𝑦))
Finding F(𝝀𝒙 ,𝝀𝒚)
F(𝜆𝑥 ,𝜆𝑦)=(𝜆𝑥 + 2(𝜆𝑦))/(𝜆𝑥 −𝜆𝑦)
=𝜆(𝑥 + 2𝑦)/(𝜆 (𝑥 − 𝑦) )
=((𝑥 + 2𝑦))/(𝑥 − 𝑦)
= F(𝑥 , 𝑦)
Thus , F(𝜆𝑥 ,𝜆𝑦)="F" (𝑥 , 𝑦)" "
=𝜆°" F" (𝑥 , 𝑦)" "
Thus , "F" (𝑥 , 𝑦)" is Homogeneous function of degree zero"
Therefore, the given Differential Equation is Homogeneous differential Equation
Step 3: Solving 𝑑𝑦/𝑑𝑥 by Putting 𝑦=𝑣𝑥
𝑑𝑦/𝑑𝑥=((𝑥 + 2𝑦)/(𝑥 − 𝑦))
Let 𝑦=𝑣𝑥
So , 𝑑𝑦/𝑑𝑥=𝑑(𝑣𝑥)/𝑑𝑥
𝑑𝑦/𝑑𝑥=𝑑𝑣/𝑑𝑥 . 𝑥+𝑣 𝑑𝑥/𝑑𝑥
𝑑𝑦/𝑑𝑥=𝑑𝑣/𝑑𝑥 𝑥+𝑣
Putting 𝑑𝑦/𝑑𝑥 𝑎nd 𝑦/𝑥 𝑖𝑛 (𝑖) 𝑖.𝑒.
𝑑𝑦/𝑑𝑥=(𝑥 + 2𝑦)/(𝑥 − 𝑦)
𝑑𝑣/𝑑𝑥 𝑥+𝑣= (𝑥 + 2 𝑣𝑥)/(𝑥 − 𝑣𝑥)
𝑑𝑣/𝑑𝑥 𝑥+𝑣= 𝑥(1 + 2𝑣)/𝑥(1 − 𝑣)
𝑑𝑣/𝑑𝑥 𝑥+𝑣= (1 + 2𝑣)/(1− 𝑣)
𝑑𝑣/𝑑𝑥 𝑥= (1 + 2𝑣)/(1− 𝑣)−𝑣
𝑑𝑣/𝑑𝑥 . 𝑥= (1 + 2𝑣 − 𝑣 (1− 𝑣))/(1 −𝑣)
𝑑𝑣/𝑑𝑥 . 𝑥= (1 + 2𝑣 − 𝑣 +〖 𝑣〗^2)/(1 − 𝑣)
𝑑𝑣/𝑑𝑥 . 𝑥= (〖 𝑣〗^2 + 𝑣 + 1)/(1 − 𝑣)
𝑑𝑣/𝑑𝑥 𝑥=−((〖 𝑣〗^2 + 𝑣 + 1)/(𝑣 − 1))
𝑑𝑣((𝑣−1)/(𝑣^(2 )+ 𝑣 + 1))=(− 𝑑𝑥)/𝑥
Integrating Both Sides
∫1▒〖(𝑣 −1)/(𝑣^2 + 𝑣 + 1) 𝑑𝑣=∫1▒(−𝑑𝑥)/𝑥〗
∫1▒〖(𝑣 −1)/(𝑣^2 + 𝑣 + 1) 𝑑𝑣=−∫1▒𝑑𝑥/𝑥〗
∫1▒〖((𝑣 −1) 𝑑𝑣)/(𝑣^2 + 𝑣 + 1) 𝑑𝑣〗=−log〖|𝑥|〗 + 𝑐
We can write
𝑣^2+𝑣+1 = 𝑣^2 + 1/2 . 2v + (1/2)^2+1−(1/2)^2
=(𝑣+1/2)^2 + 1 – 1/4
=(𝑣+1/2)^2+3/4
Putting 𝑣^2+𝑣+1=(𝑣+1/2)^2+3/4
& 𝑣−1=𝑣+𝟏/𝟐−𝟏/𝟐−1 =(𝑣+1/2)−3/2
∫1▒((𝑣 + 1/2) − 3/2)/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣=−log〖𝑥+𝑐〗
∫1▒(𝑣 + 1/2)/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣−3/2 ∫1▒1/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣=−log〖𝑥+𝑐〗
Thus, I = I1 – (3 )/2 I2
Solving 𝐼1
𝐼1=∫1▒((𝑣 + 1/2))/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣
Put (𝑣+ 1/2)^2+ 3/4 =𝑡
Diff. w.r.t. 𝑣
𝑑((𝑣 + 1/2)^2+ 3/4)/𝑑𝑣=𝑑𝑡/𝑑𝑣
2(𝑣+1/2)=𝑑𝑡/𝑑𝑣
𝑑𝑣=𝑑𝑡/2(𝑣 + 1/2)
Putting value of v & dv in I1
𝐼1=∫1▒((𝑣 + 1/2))/𝑡 ×𝑑𝑡/2(𝑣 + 1/2)
=1/2 ∫1▒𝑑𝑡/𝑡
=1/2 log |𝑡|
Putting 𝑡=(𝑣+ 1/2)^2+3/4
=1/2 𝑙𝑜𝑔|(𝑣+ 1/2)^2+3/4|
=1/2 𝑙𝑜𝑔|𝑣^2+2𝑣 × 1/2 + 1/4 + 3/4|
I1 =1/2 log〖 |𝑣^2+𝑣+1|〗
Solving 𝑰𝟐
𝐼2=∫1▒𝑑𝑣/((𝑣 + 1/2)^2+3/4)
=∫1▒𝑑𝑣/((𝑣 + 1/2)^2+(√3/2)^2 )
Put 𝑡=𝑣+1/2
Diff. w.r.t. 𝑣
𝑑𝑡/𝑑𝑣=1 ⇒ 𝑑𝑡=𝑑𝑣
= ∫1▒𝑑𝑡/(𝑡^2 + 〖 (√3/2)〗^2 )
=1/(√3/2) tan^(−1)〖𝑡/(√3/2)〗
=2/√3 tan^(−1)〖2(𝑣 + 1/2)/√3〗
=2/√3 tan^(−1)((2𝑣 + 1)/√3)
Hence
I = 𝐼1−3/2 𝐼2
I =1/2 log |𝑣^2+𝑣+1|−3/2 ×2/√3 tan^(−1)((2𝑣 +1)/√3)
I =1/2 log |𝑣^2+𝑣+1|−√3 tan^(−1)((2𝑣 +1)/√3)
Replacing v by (𝑦 )/𝑥
I =1/2 𝑙𝑜𝑔|(𝑦/𝑥)^2+𝑦/𝑥+1|−√3 tan^(−1)((2𝑦/𝑥 + 1)/√3)
I =1/2 𝑙𝑜𝑔|𝑦^2/𝑥^2 +𝑦/𝑥+1|−√3 tan^(−1)((2𝑦 + 𝑥)/(√3 𝑥))
Putting Value of I in (2)
1/2 𝑙𝑜𝑔|𝑦^2/𝑥^2 +𝑦/𝑥+1|−√3 tan^(−1)((2𝑦 + 𝑥)/(√3 𝑥))=−𝑙𝑜𝑔|𝑥|+𝑐
1/2 𝑙𝑜𝑔|𝑦^2/𝑥^2 +𝑦/𝑥+1|+𝑙𝑜𝑔|𝑥|=√3 tan^(−1)((2𝑦 + 𝑥)/(√3 𝑥))+𝑐
Multiplying Both Sides By 2
𝑙𝑜𝑔|𝑦^2/𝑥^2 +𝑦/𝑥+1|+2 𝑙𝑜𝑔|𝑥|=2 √3 tan^(−1)((2𝑦 + 𝑥)/(√3 𝑥))+2𝑐
𝑙𝑜𝑔|𝑦^2/𝑥^2 +𝑦/𝑥+1|+𝑙𝑜𝑔|𝑥|^2=2√3 tan^(−1)((2𝑦 + 𝑥)/(√3 𝑥))+2𝑐
Put 2𝑐=𝑐
𝑙𝑜𝑔[|𝑦^2/𝑥^2 +𝑦/𝑥+1| × |𝑥^2 |]=2√3 tan^(−1)((2𝑦 + 𝑥)/(√3 𝑥))+𝑐
𝑙𝑜𝑔|〖𝑥^2 𝑦〗^2/𝑥^2 +(𝑥^2 𝑦)/𝑥+𝑥^2 |=2√3 tan^(−1)((𝑥 + 2𝑦)/(√3 𝑥))+𝑐
𝒍𝒐𝒈|𝒙^𝟐+𝒙𝒚+𝒚^𝟐 |=𝟐√𝟑 〖𝒕𝒂𝒏〗^(−𝟏)((𝒙 + 𝟐𝒚)/(√𝟑 𝒙))+𝒄
Is the General Solution of the Differential Equation given

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