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 March 16, 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 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
β΄ 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¦β2π₯/π¦ = π¦^2/π^π¦
ππ₯/ππ¦+((β2)/π¦)π₯ = π¦^2/π^π¦
Comparing with ππ₯/ππ¦ + P1 x = Q1
Where P1 = (β2)/π¦ & Q1 = π¦^2/π^π¦
Find Integrating factor,
IF = π^β«1βγπ1 ππ¦γ
= π^β«1β(β2ππ¦)/π¦
= π^(β2β«1βππ¦/π¦)
= π^(β2 logβ‘π¦ )
Using a log b = log ba
= π^(γlogβ‘π¦γ^(β2) )
= π¦^(β2)
= 1/π¦^2
Solution is
π₯ (IF) = β«1βγ(π1ΓπΌπΉ)ππ¦+πγ
π₯ Γ 1/π¦^2 = β«1βγπ¦^2/π^π¦ Γ1/π¦^2 ππ¦+πγ
π₯/π¦^2 = β«1βγ1/π^π¦ ππ¦+πγ
π₯/π¦^2 = β«1βγπ^(βπ¦) ππ¦+πγ
π₯/π¦^2 = β1 Γπ^(βπ¦)+π
π₯/π¦^2 = γβπγ^(βπ¦)+π
π₯ = γβπ¦^2 πγ^(βπ¦)+ππ¦^2
We need to find particular solution,
Putting x = 0, y = 1
0 = γβ1^2 πγ^(β1)+πγ(1)γ^2
0 = γβπγ^(β1)+π
π^(β1) =π
c=π^(β1)
c=1/π
Putting value of c in (2)
π₯ = γβπ¦^2 πγ^(βπ¦)+ππ¦^2
π₯ = γβπ¦^2 πγ^(βπ¦)+(1/π) π¦^2
π = γβπ^π πγ^(βπ)+π^π/π
Question 19
Show that (x β 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 β y)dy = (x + 2y)dx
ππ¦/ππ₯=((π₯ + 2π¦)/(π₯ β π¦))
Now, Putting F(π₯ , π¦)=ππ¦/ππ₯ & Find F(ππ₯ ,ππ¦)
Let F(π₯ , π¦)=((π₯ + 2π¦)/(π₯ β π¦))
Finding F(ππ ,ππ)
F(ππ₯ ,ππ¦)=(ππ₯ + 2(ππ¦))/(ππ₯ βππ¦)
=π(π₯ + 2π¦)/(π (π₯ β π¦) )
=((π₯ + 2π¦))/(π₯ β π¦)
= F(π₯ , π¦)
Thus , F(ππ₯ ,ππ¦)="F" (π₯ , π¦)" "
=πΒ°" 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₯ β 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₯ β π¦)
ππ£/ππ₯ π₯+π£= (π₯ + 2 π£π₯)/(π₯ β π£π₯)
ππ£/ππ₯ π₯+π£= π₯(1 + 2π£)/π₯(1 β π£)
ππ£/ππ₯ π₯+π£= (1 + 2π£)/(1β π£)
ππ£/ππ₯ π₯= (1 + 2π£)/(1β π£)βπ£
ππ£/ππ₯ . π₯= (1 + 2π£ β π£ (1β π£))/(1 βπ£)
ππ£/ππ₯ . π₯= (1 + 2π£ β π£ +γ π£γ^2)/(1 β π£)
ππ£/ππ₯ . π₯= (γ π£γ^2 + π£ + 1)/(1 β π£)
ππ£/ππ₯ π₯=β((γ π£γ^2 + π£ + 1)/(π£ β 1))
ππ£((π£β1)/(π£^(2 )+ π£ + 1))=(β ππ₯)/π₯
Integrating Both Sides
β«1βγ(π£ β1)/(π£^2 + π£ + 1) ππ£=β«1β(βππ₯)/π₯γ
β«1βγ(π£ β1)/(π£^2 + π£ + 1) ππ£=ββ«1βππ₯/π₯γ
β«1βγ((π£ β1) ππ£)/(π£^2 + π£ + 1) ππ£γ=βlogβ‘γ|π₯|γ + π
We can write
π£^2+π£+1 = π£^2 + 1/2 . 2v + (1/2)^2+1β(1/2)^2
=(π£+1/2)^2 + 1 β 1/4
=(π£+1/2)^2+3/4
Putting π£^2+π£+1=(π£+1/2)^2+3/4
& π£β1=π£+π/πβπ/πβ1 =(π£+1/2)β3/2
β«1β((π£ + 1/2) β 3/2)/((π£ + 1/2)^2+ 3/4) ππ£=βlogβ‘γπ₯+πγ
β«1β(π£ + 1/2)/((π£ + 1/2)^2+ 3/4) ππ£β3/2 β«1β1/((π£ + 1/2)^2+ 3/4) ππ£=βlogβ‘γπ₯+πγ
Thus, I = I1 β (3 )/2 I2
Solving πΌ1
πΌ1=β«1β((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=β«1β((π£ + 1/2))/π‘ Γππ‘/2(π£ + 1/2)
=1/2 β«1βππ‘/π‘
=1/2 logβ‘ |π‘|
Putting π‘=(π£+ 1/2)^2+3/4
=1/2 πππ|(π£+ 1/2)^2+3/4|
=1/2 πππ|π£^2+2π£ Γ 1/2 + 1/4 + 3/4|
I1 =1/2 logβ‘γ |π£^2+π£+1|γ
Solving π°π
πΌ2=β«1βππ£/((π£ + 1/2)^2+3/4)
=β«1βππ£/((π£ + 1/2)^2+(β3/2)^2 )
Put π‘=π£+1/2
Diff. w.r.t. π£
ππ‘/ππ£=1 β ππ‘=ππ£
= β«1βππ‘/(π‘^2 + γ (β3/2)γ^2 )
=1/(β3/2) tan^(β1)β‘γπ‘/(β3/2)γ
=2/β3 tan^(β1)β‘γ2(π£ + 1/2)/β3γ
=2/β3 tan^(β1)β‘((2π£ + 1)/β3)
Hence
I = πΌ1β3/2 πΌ2
I =1/2 logβ‘ |π£^2+π£+1|β3/2 Γ2/β3 tan^(β1)β‘((2π£ +1)/β3)
I =1/2 logβ‘ |π£^2+π£+1|ββ3 tan^(β1)β‘((2π£ +1)/β3)
Replacing v by (π¦ )/π₯
I =1/2 πππ|(π¦/π₯)^2+π¦/π₯+1|ββ3 tan^(β1)β‘((2π¦/π₯ + 1)/β3)
I =1/2 πππ|π¦^2/π₯^2 +π¦/π₯+1|ββ3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))
Putting Value of I in (2)
1/2 πππ|π¦^2/π₯^2 +π¦/π₯+1|ββ3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))=βπππ|π₯|+π
1/2 πππ|π¦^2/π₯^2 +π¦/π₯+1|+πππ|π₯|=β3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))+π
Multiplying Both Sides By 2
πππ|π¦^2/π₯^2 +π¦/π₯+1|+2 πππ|π₯|=2 β3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))+2π
πππ|π¦^2/π₯^2 +π¦/π₯+1|+πππ|π₯|^2=2β3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))+2π
Put 2π=π
πππ[|π¦^2/π₯^2 +π¦/π₯+1| Γ |π₯^2 |]=2β3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))+π
πππ|γπ₯^2 π¦γ^2/π₯^2 +(π₯^2 π¦)/π₯+π₯^2 |=2β3 tan^(β1)β‘((π₯ + 2π¦)/(β3 π₯))+π
πππ|π^π+ππ+π^π |=πβπ γπππγ^(βπ)β‘((π + ππ)/(βπ π))+π
Is the General Solution of the Differential Equation given
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