JEE 2023 — Mathematics PYQ
JEE | Mathematics | 2023Let the solution curve y=y(x) of the differential equation
pass through the origin. Then y(1) is equal to:
Choose the correct answer:
- A.
exp(424+π)
- B.
exp(421−π)
exp(424−π)
Explanation
Solution
Integrating Factor (I.F.):
I.F.=e∫P(x)dx=exp(∫−(1+x6)3/23x5tan−1(x3)dx)
Let x3=tanθ⟹3x2dx=sec2θdθ:
∫−sec3θtanθ⋅θ⋅sec2θdθ=∫−θsinθdθ=θcosθ−sinθ
I.F.=exp(1+x6tan−1x3−1+x6x3)=exp(1+x6tan−1x3−x3)
Solution of D.E.:
y⋅e1+x6tan−1x3−x3=∫2x⋅e1+x6x3−tan−1x3⋅e1+x6tan−1x3−x3dx
y⋅e1+x6tan−1x3−x3=∫2xdx=x2+c
As it passes through (0,0)⟹c=0:
y=x2exp(1+x6x3−tan−1x3)
For x=1:
y(1)=12⋅exp(1+11−tan−11)=exp(21−π/4)=exp(424−π)
Correct Option: (4)
Explanation
Solution
Integrating Factor (I.F.):
I.F.=e∫P(x)dx=exp(∫−(1+x6)3/23x5tan−1(x3)dx)
Let x3=tanθ⟹3x2dx=sec2θdθ:
∫−sec3θtanθ⋅θ⋅sec2θdθ=∫−θsinθdθ=θcosθ−sinθ
I.F.=exp(1+x6tan−1x3−1+x6x3)=exp(1+x6tan−1x3−x3)
Solution of D.E.:
y⋅e1+x6tan−1x3−x3=∫2x⋅e1+x6x3−tan−1x3⋅e1+x6tan−1x3−x3dx
y⋅e1+x6tan−1x3−x3=∫2xdx=x2+c
As it passes through (0,0)⟹c=0:
y=x2exp(1+x6x3−tan−1x3)
For x=1:
y(1)=12⋅exp(1+11−tan−11)=exp(21−π/4)=exp(424−π)
Correct Option: (4)

