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Differential Equations question

2023 · 30 Jan · Shift 1 · Q23
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  5. /2023 · 30 Jan · Shift 1 · Q23

Differential Equations question

2023 · 30 Jan · Shift 1 · Q23

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let the solution curve y=y(x)y=y(x)y=y(x) of the differential equation dy dx−3x5tan⁡−1(x3)(1+x6)3/2y=2xexp⁡{x3−tan⁡−1x3(1+x6)} pass through the origin. Then y(1) is equal to : \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\} \text { pass through the origin. Then } y(1) \text { is equal to : } dxdy​−(1+x6)3/23x5tan−1(x3)​y=2xexp{(1+x6)​x3−tan−1x3​} pass through the origin. Then y(1) is equal to : 
  1. A
    exp⁡(1−π42)\exp \left(\frac{1-\pi}{4 \sqrt{2}}\right)exp(42​1−π​)
  2. B
    exp⁡(4−π42)\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right)exp(42​4−π​)
  3. C
    exp⁡(4+π42)\exp \left(\frac{4+\pi}{4 \sqrt{2}}\right)exp(42​4+π​)
  4. D
    exp⁡(π−442)\exp \left(\frac{\pi-4}{4 \sqrt{2}}\right)exp(42​π−4​)
View written solutionFree

Correct answer: B

  1. Write the differential equation in linear form

Given

dydx−3x5tan⁡−1(x3)(1+x6)3/2y=2xexp⁡{x3−tan⁡−1(x3)1+x6}.\frac{dy}{dx}-\frac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}y=2x\exp\left\{\frac{x^3-\tan^{-1}(x^3)}{\sqrt{1+x^6}}\right\}.dxdy​−(1+x6)3/23x5tan−1(x3)​y=2xexp{1+x6​x3−tan−1(x3)​}.

This is a linear differential equation of the form

dydx+P(x)y=Q(x),\frac{dy}{dx}+P(x)y=Q(x),dxdy​+P(x)y=Q(x),

with

P(x)=−3x5tan⁡−1(x3)(1+x6)3/2,Q(x)=2xexp⁡{x3−tan⁡−1(x3)1+x6}.P(x)=-\frac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}, \qquad Q(x)=2x\exp\left\{\frac{x^3-\tan^{-1}(x^3)}{\sqrt{1+x^6}}\right\}.P(x)=−(1+x6)3/23x5tan−1(x3)​,Q(x)=2xexp{1+x6​x3−tan−1(x3)​}.
  1. Find the integrating factor

We need

I.F.=e∫P(x) dx=e∫−3x5tan⁡−1(x3)(1+x6)3/2 dx.\text{I.F.}=e^{\int P(x)\,dx} =e^{\int -\frac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}\,dx}.I.F.=e∫P(x)dx=e∫−(1+x6)3/23x5tan−1(x3)​dx.

Now observe that

ddx(tan⁡−1(x3)1+x6)=3x21+x6⋅11+x6+tan⁡−1(x3)⋅ddx(1+x6)−1/2.\frac{d}{dx}\left(\frac{\tan^{-1}(x^3)}{\sqrt{1+x^6}}\right) = \frac{3x^2}{1+x^6}\cdot \frac{1}{\sqrt{1+x^6}} +\tan^{-1}(x^3)\cdot \frac{d}{dx}(1+x^6)^{-1/2}.dxd​(1+x6​tan−1(x3)​)=1+x63x2​⋅1+x6​1​+tan−1(x3)⋅dxd​(1+x6)−1/2.

But the useful quantity here is

ddx(x3−tan⁡−1(x3)1+x6).\frac{d}{dx}\left(\frac{x^3-\tan^{-1}(x^3)}{\sqrt{1+x^6}}\right).dxd​(1+x6​x3−tan−1(x3)​).

Let

ϕ(x)=x3−tan⁡−1(x3)1+x6.\phi(x)=\frac{x^3-\tan^{-1}(x^3)}{\sqrt{1+x^6}}.ϕ(x)=1+x6​x3−tan−1(x3)​.

Differentiate:

ddx(x3−tan⁡−1(x3))=3x2−3x21+x6=3x81+x6.\frac{d}{dx}(x^3-\tan^{-1}(x^3)) =3x^2-\frac{3x^2}{1+x^6} =\frac{3x^8}{1+x^6}.dxd​(x3−tan−1(x3))=3x2−1+x63x2​=1+x63x8​.

Also,

ddx(1+x6)−1/2=−12(1+x6)−3/2(6x5)=−3x5(1+x6)3/2.\frac{d}{dx}(1+x^6)^{-1/2}=-\frac{1}{2}(1+x^6)^{-3/2}(6x^5)=-\frac{3x^5}{(1+x^6)^{3/2}}.dxd​(1+x6)−1/2=−21​(1+x6)−3/2(6x5)=−(1+x6)3/23x5​.

So

ϕ′(x)=3x8(1+x6)3/2−(x3−tan⁡−1(x3))3x5(1+x6)3/2.\phi'(x)=\frac{3x^8}{(1+x^6)^{3/2}}-(x^3-\tan^{-1}(x^3))\frac{3x^5}{(1+x^6)^{3/2}}.ϕ′(x)=(1+x6)3/23x8​−(x3−tan−1(x3))(1+x6)3/23x5​.

Simplifying,

ϕ′(x)=3x8−3x8+3x5tan⁡−1(x3)(1+x6)3/2=3x5tan⁡−1(x3)(1+x6)3/2.\phi'(x)=\frac{3x^8-3x^8+3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}} =\frac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}.ϕ′(x)=(1+x6)3/23x8−3x8+3x5tan−1(x3)​=(1+x6)3/23x5tan−1(x3)​.

Hence

P(x)=−ϕ′(x).P(x)=-\phi'(x).P(x)=−ϕ′(x).

Therefore the integrating factor is

I.F.=e∫P(x)dx=e−ϕ(x)=exp⁡(−x3−tan⁡−1(x3)1+x6).\text{I.F.}=e^{\int P(x)dx}=e^{-\phi(x)} =\exp\left(-\frac{x^3-\tan^{-1}(x^3)}{\sqrt{1+x^6}}\right).I.F.=e∫P(x)dx=e−ϕ(x)=exp(−1+x6​x3−tan−1(x3)​).
  1. Multiply the equation by the integrating factor

Then

ddx(y e−ϕ(x))=Q(x)e−ϕ(x).\frac{d}{dx}\big(y\,e^{-\phi(x)}\big)=Q(x)e^{-\phi(x)}.dxd​(ye−ϕ(x))=Q(x)e−ϕ(x).

Since

Q(x)=2xeϕ(x),Q(x)=2x e^{\phi(x)},Q(x)=2xeϕ(x),

we get

Q(x)e−ϕ(x)=2x.Q(x)e^{-\phi(x)}=2x.Q(x)e−ϕ(x)=2x.

Therefore

ddx(ye−ϕ(x))=2x.\frac{d}{dx}\big(y e^{-\phi(x)}\big)=2x.dxd​(ye−ϕ(x))=2x.

Integrating,

ye−ϕ(x)=x2+C.ye^{-\phi(x)}=x^2+C.ye−ϕ(x)=x2+C.

Thus

y=(x2+C)eϕ(x).y=(x^2+C)e^{\phi(x)}.y=(x2+C)eϕ(x).
  1. Use the condition that the curve passes through the origin

Passing through the origin means

y(0)=0.y(0)=0.y(0)=0.

Now

ϕ(0)=0−tan⁡−1(0)1=0.\phi(0)=\frac{0-\tan^{-1}(0)}{\sqrt{1}}=0.ϕ(0)=1​0−tan−1(0)​=0.

So

y(0)=(0+C)e0=C.y(0)=(0+C)e^0=C.y(0)=(0+C)e0=C.

Hence

C=0.C=0.C=0.

Therefore

y=x2exp⁡(x3−tan⁡−1(x3)1+x6).y=x^2\exp\left(\frac{x^3-\tan^{-1}(x^3)}{\sqrt{1+x^6}}\right).y=x2exp(1+x6​x3−tan−1(x3)​).
  1. Find y(1)y(1)y(1)

Substitute x=1x=1x=1:

y(1)=12exp⁡(1−tan⁡−1(1)2).y(1)=1^2\exp\left(\frac{1-\tan^{-1}(1)}{\sqrt{2}}\right).y(1)=12exp(2​1−tan−1(1)​).

Since

tan⁡−1(1)=π4,\tan^{-1}(1)=\frac{\pi}{4},tan−1(1)=4π​,

we get

y(1)=exp⁡(1−π/42)=exp⁡(4−π42).y(1)=\exp\left(\frac{1-\pi/4}{\sqrt{2}}\right) =\exp\left(\frac{4-\pi}{4\sqrt{2}}\right).y(1)=exp(2​1−π/4​)=exp(42​4−π​).
  1. Compare with the options

This matches

B: exp⁡(4−π42).\boxed{\text{B: } \exp\left(\frac{4-\pi}{4\sqrt{2}}\right)}.B: exp(42​4−π​)​.
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