If y = y(x) is the solution curve of the differential equation ( x 2 - 4 ) d y - ( y 2 - 3 y ) d x = 0 , x > 2 , y ( 4 ) = 3 2 and the slope of the curve is never zero, then the value of y(10) equals: [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
If y = y(x) is the solution curve of the differential equation $(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0, x > 2, y (4) = \frac{3}{2}$ and the slope of the curve is never zero, then the value of y(10) equals: [2024]

1 $\frac{3}{1 + {(8)}^{1 / 4}}$

2 $\frac{3}{1 - {(8)}^{1 / 4}}$

3 $\frac{3}{1 - 2 \sqrt{2}}$

4 $\frac{3}{1 + 2 \sqrt{2}}$

Ans.
(1)

Given differential equation is

$(x^{2} - 4) d y - (y^{2} - 3 y) d x = 0 \Rightarrow \frac{d y}{y^{2} - 3 y} = \frac{d x}{x^{2} - 4}$

Integrating on both sides, we get

$\int \frac{d y}{y^{2} - 3 y} = \int \frac{d x}{x^{2} - 4}$

$\Rightarrow \frac{1}{3} \int \frac{y - (y - 3)}{y (y - 3)} d y = \frac{1}{4} \int \frac{(x + 2) - (x - 2)}{(x - 2) (x + 2)} d x$

$\Rightarrow \int \frac{d y}{3 (y - 3)} - \int \frac{d y}{3 y} = \int \frac{d x}{4 (x - 2)} - \int \frac{d x}{4 (x + 2)}$

$\Rightarrow \frac{1}{3} \ln | \frac{y - 3}{y} | = \frac{1}{4} \ln | \frac{x - 2}{x + 2} | + \ln C \Rightarrow \frac{y - 3}{y} = \frac{C_{1} {(x - 2)}^{3 / 4}}{{(x + 2)}^{3 / 4}}$

$\Rightarrow y = \frac{3 {(x + 2)}^{3 / 4}}{{(x + 2)}^{3 / 4} - C_{1} {(x - 2)}^{3 / 4}}$

Now, $y (4) = \frac{3}{2} \Rightarrow C_{1} = - 3^{3 / 4}$

$∴ y = \frac{3 {(x + 2)}^{3 / 4}}{{(x + 2)}^{3 / 4} + {(3 x - 6)}^{3 / 4}}$

So, $y (10) = \frac{3 \times 12^{3 / 4}}{12^{3 / 4} + 24^{3 / 4}} = \frac{3}{1 + 2^{3 / 4}} = \frac{3}{1 + {(8)}^{1 / 4}}$

Want Us to Email you About Special Offers & Updates?