Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • If the solution of the differential equation <img alt="\mathrm{\frac{d y}{d x}+e^{x}\left(x^{2}-2\right) y}=\mathrm{\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x}}" src="https://entrance

If the solution of the differential equation
\mathrm{\frac{d y}{d x}+e^{x}\left(x^{2}-2\right) y}=\mathrm{\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x}} satisfies \mathrm{y(0)=0}, then the value of \mathrm{y(2)} is_______

Option: 1

-1


Option: 2

10


Option: 3

0


Option: 4

e


Answers (1)

best_answer

\begin{aligned} &\frac{d y}{d x}+e^{x}\left(x^{2}-2\right) y=\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x}\\ &\begin{aligned} I \cdot F \cdot &=e^{\int e^{x}\left(x^{2}-2\right) d x} \\ &=e^{\left[\left(x^{2}-2\right) e^{x}-\int 2 x e^{x} d x\right]} \end{aligned}\\ \end{aligned}

\begin{aligned}&\begin{aligned} &=e^{\left[\left(x^{2}-2\right) e^{x}-2 e^{x}(x)+2 e^{x}\right]} \end{aligned}\\ \end{aligned}\begin{aligned}&=e^{\left(x^{2}-2 x\right) e^{x}}\\ &y \cdot e^{\left(x^{2}-2 x\right) e^{x}}=\int e^{\left(x^{2}-2 x\right) e^{x}}\left(x^{2}-2 x\right)\left(x^{2}-2\right) e^{2 x} d x\\ &e^{x}\left(x^{2}-2 x\right)=t\\ &\left\{2(x-1) e^{x}+e^{x}\left(x^{2}-2 x\right)\right\} d x=d x\\ &y \cdot e^{t}=\int e^{t} \cdot t d t\\ &y e^{t}=t e^{t}-e^{t}+c\\ &y e^{\left(x^{2}-2 x\right) e^{x}}=\left(x^{2}-2 x\right) e^{x} e^{\left(x^{2}-2 x\right) e^{x}}-e^{\left(x^{2}-2 x\right) e^{x}}+c \end{aligned}

\begin{aligned} &\text { at } x=0, y=0 \\ &0=0-1+c \Rightarrow c=\frac{1}{=} \\ &\therefore \text { at } x=2 \\ &y(1)=0-1+1 \\ &y=0 \end{aligned}

Posted by

admin

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 ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question