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Differential Equations question

2023 · 15 Apr · Shift 1 · Q24
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  5. /2023 · 15 Apr · Shift 1 · Q24

Differential Equations question

2023 · 15 Apr · Shift 1 · Q24

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let x=x(y)x=x(y)x=x(y) be the solution of the differential equation 2(y+2)log⁡e(y+2)dx+(x+4−2log⁡e(y+2))dy=0,y>−12(y+2) \log _{e}(y+2) d x+\left(x+4-2 \log _{e}(y+2)\right) d y=0, y\gt -12(y+2)loge​(y+2)dx+(x+4−2loge​(y+2))dy=0,y>−1 with x(e4−2)=1x\left(e^{4}-2\right)=1x(e4−2)=1. Then x(e9−2)x\left(e^{9}-2\right)x(e9−2) is equal to :
  1. A
    49\frac{4}{9}94​
  2. B
    329\frac{32}{9}932​
  3. C
    103\frac{10}{3}310​
  4. D
    3
View written solutionFree

Correct answer: B

  1. Rewrite the differential equation in terms of dxdy\dfrac{dx}{dy}dydx​.

Given 2(y+2)ln⁡(y+2) dx+(x+4−2ln⁡(y+2))dy=0.2(y+2)\ln(y+2)\,dx+\left(x+4-2\ln(y+2)\right)dy=0.2(y+2)ln(y+2)dx+(x+4−2ln(y+2))dy=0. Since x=x(y)x=x(y)x=x(y), divide by dydydy: 2(y+2)ln⁡(y+2)dxdy+x+4−2ln⁡(y+2)=0.2(y+2)\ln(y+2)\frac{dx}{dy}+x+4-2\ln(y+2)=0.2(y+2)ln(y+2)dydx​+x+4−2ln(y+2)=0. So, dxdy+12(y+2)ln⁡(y+2)x=2ln⁡(y+2)−42(y+2)ln⁡(y+2).\frac{dx}{dy}+\frac{1}{2(y+2)\ln(y+2)}x=\frac{2\ln(y+2)-4}{2(y+2)\ln(y+2)}.dydx​+2(y+2)ln(y+2)1​x=2(y+2)ln(y+2)2ln(y+2)−4​. That is, dxdy+12(y+2)ln⁡(y+2)x=ln⁡(y+2)−2(y+2)ln⁡(y+2).\frac{dx}{dy}+\frac{1}{2(y+2)\ln(y+2)}x=\frac{\ln(y+2)-2}{(y+2)\ln(y+2)}.dydx​+2(y+2)ln(y+2)1​x=(y+2)ln(y+2)ln(y+2)−2​.

  1. This is a linear differential equation dxdy+P(y)x=Q(y),\frac{dx}{dy}+P(y)x=Q(y),dydx​+P(y)x=Q(y), where P(y)=12(y+2)ln⁡(y+2).P(y)=\frac{1}{2(y+2)\ln(y+2)}.P(y)=2(y+2)ln(y+2)1​.

  2. Find the integrating factor.

IF=e∫P(y) dy=e∫12(y+2)ln⁡(y+2)dy.IF=e^{\int P(y)\,dy}=e^{\int \frac{1}{2(y+2)\ln(y+2)}dy}.IF=e∫P(y)dy=e∫2(y+2)ln(y+2)1​dy. Let t=ln⁡(y+2)  ⟹  dt=dyy+2.t=\ln(y+2)\implies dt=\frac{dy}{y+2}.t=ln(y+2)⟹dt=y+2dy​. Then ∫12(y+2)ln⁡(y+2)dy=12∫1tdt=12ln⁡t.\int \frac{1}{2(y+2)\ln(y+2)}dy=\frac12\int \frac{1}{t}dt=\frac12\ln t.∫2(y+2)ln(y+2)1​dy=21​∫t1​dt=21​lnt. Hence IF=e12ln⁡(ln⁡(y+2))=ln⁡(y+2).IF=e^{\frac12\ln(\ln(y+2))}=\sqrt{\ln(y+2)}.IF=e21​ln(ln(y+2))=ln(y+2)​.

  1. Multiply the equation by the integrating factor.
=\frac{\ln(y+2)-2}{(y+2)\sqrt{\ln(y+2)}}.$$ Left side is $$\frac{d}{dy}\left(x\sqrt{\ln(y+2)}\right).$$ Therefore, $$\frac{d}{dy}\left(x\sqrt{\ln(y+2)}\right)=\frac{\ln(y+2)-2}{(y+2)\sqrt{\ln(y+2)}}.$$ 5. **Integrate both sides.** Again put $$u=\ln(y+2),\qquad du=\frac{dy}{y+2}.$$ Then $$\int \frac{\ln(y+2)-2}{(y+2)\sqrt{\ln(y+2)}}dy =\int \frac{u-2}{\sqrt{u}}du =\int \left(\sqrt{u}-2u^{-1/2}\right)du.$$ So, $$\int \left(\sqrt{u}-2u^{-1/2}\right)du =\frac{2}{3}u^{3/2}-4u^{1/2}+C.$$ Thus, $$x\sqrt{\ln(y+2)}=\frac{2}{3}(\ln(y+2))^{3/2}-4(\ln(y+2))^{1/2}+C.$$ Divide by $\sqrt{\ln(y+2)}$: $$x=\frac{2}{3}\ln(y+2)-4+\frac{C}{\sqrt{\ln(y+2)}}.$$ 6. **Use the initial condition** $x(e^4-2)=1$. When $$y=e^4-2,\quad y+2=e^4,\quad \ln(y+2)=4.$$ Hence $$1=\frac{2}{3}(4)-4+\frac{C}{\sqrt{4}}= rac{8}{3}-4+\frac{C}{2}.$$ Now $$\frac{8}{3}-4=-\frac{4}{3},$$ so $$1=-\frac{4}{3}+\frac{C}{2}$$ $$\frac{C}{2}=1+\frac{4}{3}=\frac{7}{3}$$ $$C=\frac{14}{3}.$$ Therefore, $$x(y)=\frac{2}{3}\ln(y+2)-4+\frac{14}{3\sqrt{\ln(y+2)}}.$$ 7. **Find** $x(e^9-2)$. When $$y=e^9-2,\quad y+2=e^9,\quad \ln(y+2)=9,\quad \sqrt{\ln(y+2)}=3.$$ So, $$x(e^9-2)=\frac{2}{3}(9)-4+\frac{14}{3\cdot 3}$$ $$=6-4+\frac{14}{9}$$ $$=2+\frac{14}{9}=\frac{18+14}{9}=\frac{32}{9}.$$ 8. **Check options.** $$\boxed{\frac{32}{9}}$$ which is **Option B**.
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