Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve this problem - Differential equations - JEE Main-2

If  y(x) is the solution of the differential equation (x+2)\frac{dy}{dx}=x^{2}+4x-9,x\neq -2\: and\, \, y(0)=0,\; then\; y(-4)\; is\; equal\; to\, :

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    -1

  • Option 4)

    2

 

Answers (1)

As we learnt in

Solution of Differential Equation -

\frac{\mathrm{d}y }{\mathrm{d} x} =f\left ( ax+by+c \right )

put

 Z =ax+by+c

 

 

- wherein

Equation with convert to

\int \frac{dz}{bf\left ( z \right )+a} =x+c

 

 

 

 (x+2)\frac{dy}{dx}=x^{2}+4x-9

\int dy=\int \frac{x^{2}+4x-9}{x+2}dx

\int dy=\int \frac{x^{2}+4x+4-13}{x+2}dx

        =\int \frac{(x+2)^{2}}{(x+2)}dx -\int \frac{13}{x+2}dx

        =\int (x+2)dx-\int \frac{13}{x+2}dx

y=\frac{x^{2}}{2}+2x-13\ log\left | x+2 \right |+C

0=0+0-13\ log2+C

\therefore C=13\ log2

y=\frac{x^{2}}{2}+2x-13log\left | x+2 \right |+13\ log2

y=\frac{16}{2}-8-13\ log\left | -4+2 \right |+13\ log2

=8 - 8 - 1 log | 2 | + 13 log 2 = 0 

 


Option 1)

0

This option is correct 

Option 2)

1

This option is incorrect 

Option 3)

-1

This option is incorrect 

Option 4)

2

This option is incorrect 

Posted by

Vakul

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 2024 session 2 result date announced; Steps to check NTA session 2 scores
anam-khan

JEE Main 2024 session 2 final answer key soon at jeemain.nta.ac.in; result on April 25
anam-khan

JEE Main 2024 Session 2: Last day to challenge answer key today

Latest Question