Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let the solution curve y = y(x) of the differential equation <img alt="\frac{d y}{d x}-\frac{3 x^5 \tan ^{-1}\left(x^3\right)}{\left(1+x^6\right)^{\frac{3}{2}}} y=2 x \exp \left\{\frac{x^3-\ta

Let the solution curve y = y(x) of the differential equation \frac{d y}{d x}-\frac{3 x^5 \tan ^{-1}\left(x^3\right)}{\left(1+x^6\right)^{\frac{3}{2}}} y=2 x \exp \left\{\frac{x^3-\tan ^{-1} x^3}{\sqrt{\left(1+x^6\right)}}\right\}

pass through the origin. Then y(1) is equal to :

Option: 1

\exp \left(\frac{4+\pi}{4 \sqrt{2}}\right)


Option: 2

\exp \left(\frac{1-\pi}{4 \sqrt{2}}\right)


Option: 3

\exp \left(\frac{\pi-4}{4 \sqrt{2}}\right)


Option: 4

\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right)


Answers (1)

best_answer

\left(\frac{d y}{d x}\right)-\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)

above equation is linear differential equation.

\text { I.F. }=e^{\int \frac{-3 x^5 \tan ^{-1}\left(x^3\right)}{\left(1+x^6\right)^{32}} d x} \\

            =e^{-\int \frac{3 x^2 \cdot x^3 \tan ^{-1}\left(x^3\right)}{\left(1+x^6\right)^{3 / 2}} d x}

Let \tan ^{-1}\left(x^3\right)=t  then

\begin{aligned} & \frac{3 x^2 \cdot d x}{1+x^6}=d t \\ & =e^{-\int \frac{t \tan t}{\sqrt{1+\operatorname{lan}^2 t} d t}} \\ & =e^{-\int \frac{t \tan t}{\sec t} d t} \\ & =e^{-\int t \sin t d t} \\ & =e^{-[-t \cos t+\sin t]} \\ & =e^{t \operatorname{cost} t-\sin t} \end{aligned}

\text { I.F. }=e^{\frac{\tan ^{-1} x^5}{\sqrt{1+x^6}} \frac{x^3}{\sqrt{1+x^6}}}

Solution is

=\begin{aligned} & y\left(e^{\frac{\tan ^{-1} x^3-x^3}{\sqrt{1+x^6}}}\right)=\int 2 x d x=x^2+c \\ & \end{aligned}

=y\left(e^{\frac{\tan ^{-1} x^3-x^3}{\sqrt{1+x^6}}}\right)=\int 2 x d x=x^2+c

above eq. is passing through (0, 0) then c = 0

y=x^2 e^{\frac{x^3-\tan ^{-1} x^3}{\sqrt{1+x^6}}}

Put x = 1 then

\begin{aligned} &= y(1)=\mathrm{e}^{\frac{1-\frac{\pi}{4}}{\sqrt{2}}}=\mathrm{e}^{\frac{4-\pi}{4 \sqrt{2}}} \\ & \Rightarrow y(1)=\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right) \end{aligned}

 

 

 

Posted by

himanshu.meshram

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: BTech CSE cut-offs for PwD candidates at NITs make entry easy; top institutes still selective
anam-khan

73 or 99 percentile? JEE Main result questioned, J-K candidate says, ‘Don’t create hype’
anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile

Latest Question