JEE 2023 Mathematics PYQ — If is the solution of the differential equation such that and the… | Mathem Solvex | Mathem Solvex
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JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023
If y=y(x) is the solution of the differential equation \frac{dy}{dx} + \frac{4x}{(x^2 - 1)}y = \frac{x + 2}{(x^2 - 1)^{\frac{1}{2}}}, x > 1such that y(2)=92loge(2+3) and y(2)=αloge(α+β)+β−γ,α,β,γ∈N,then αβγ is equal to ________.
Choose the correct answer:
A.
6
(Correct Answer)
B.
7
C.
8
D.
9
Correct Answer:
6
Explanation
Given that
dxdy+x2−14xy=(x2−1)25x+2
which is LDE, where P=x2−14x and Q=(x2−1)5/2x+2
So I.F.=e∫Pdx=e∫x2−14xdx=e2log(x2−1)=(x2−1)2
So, solution of DE is given by
y×(x2−1)2=∫(x2−1)25x+2×(x2−1)2dx+C
=∫(x2−1)21x+2dx+C
=∫x2−1xdx+2∫x2−11dx+C
y(x2−1)2=x2−1+2log∣x+x2−1∣+C
Putting y(2)=92log(2+3)
92log(2+3)×9=3+2log∣2+3∣+C
⇒C=−3
putting x=2
y(1)2=1+2log∣2+1∣−3
⇒y=1−3+2log∣1+2∣=αlog(α+β)+β−γ
On comparing ⇒α=2,β=1,γ=3
and α⋅β⋅γ=2×1×3=6
Explanation
Given that
dxdy+x2−14xy=(x2−1)25x+2
which is LDE, where P=x2−14x and Q=(x2−1)5/2x+2
So I.F.=e∫Pdx=e∫x2−14xdx=e2log(x2−1)=(x2−1)2
So, solution of DE is given by
y×(x2−1)2=∫(x2−1)25x+2×(x2−1)2dx+C
=∫(x2−1)21x+2dx+C
=∫x2−1xdx+2∫x2−11dx+C
y(x2−1)2=x2−1+2log∣x+x2−1∣+C
Putting y(2)=92log(2+3)
92log(2+3)×9=3+2log∣2+3∣+C
⇒C=−3
putting x=2
y(1)2=1+2log∣2+1∣−3
⇒y=1−3+2log∣1+2∣=αlog(α+β)+β−γ
On comparing ⇒α=2,β=1,γ=3
and α⋅β⋅γ=2×1×3=6
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