The solution of the differential equation $y_{1} y_{3} = 3 y_{2}^{2}$ is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$x = A_{1} y + A_{2}$

$x = A_{1} y^{2} + A_{2} y$

none of these

$x = A_{1} y^{2} + A_{2} y + A_{3}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

We have,

$y_{1} y_{3} = 3 y_{2}^{2} \Rightarrow \frac{y_{3}}{y_{2}} = 3 \frac{y_{2}}{y_{1}}$

Integrating both sides, we get

$\begin{array}{l} \log y_{2} = 3 \log y_{1} + \log c_{1} \\ \Rightarrow y_{2} = c_{1} y_{1}^{3} \Rightarrow \frac{y_{2}}{y_{1}^{3}} = c_{1} \Rightarrow \frac{d y_{1}}{{y_{1}}^{3}} = c_{1} \end{array}$

Integrating both sides with respect to x, we get

$\begin{array}{l} - \frac{1}{2 y_{1}^{2}} = c_{1} x + c_{2} \\ \Rightarrow {y_{1}}^{2} = \frac{1}{- 2 c_{1} x - 2 {c_{2}}_{}} \\ \Rightarrow {y_{1}}^{2} = \frac{1}{a x + b}, where a = - 2 c_{1}, b = - 2 c_{2} \\ \Rightarrow y_{1} = \frac{1}{\sqrt{a x} + b} \end{array}$

Integrating both sides with respect to x, we get

$\begin{array}{l} y = \frac{2}{a} \sqrt{a x + b} + c_{3} \\ \Rightarrow \frac{a y - c_{3}}{2} = \sqrt{a x + b} \\ \Rightarrow a x + b = {(\frac{a y - c_{3}}{2})}^{2} \\ \Rightarrow x = \frac{a}{4} y^{2} - \frac{c_{3}}{2} y + \frac{1}{a} (\frac{c_{3}^{2}}{4} - b) \\ \Rightarrow x = A_{1} y^{2} + A_{2} y + A_{3}, \end{array}$