Download App

Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
Solve the following differential equation:
$\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2-2\text{y}^2+\text{xy}$
✓

Answer

Consider the given differential equation
$\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2-2\text{y}^2+\text{xy}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-2\text{y}^2+\text{xy}}{\text{x}^2}$
This is a homogeneous differential equation.
Substituting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we have
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2-2\text{v}^2\times\text{x}^2+\text{x}\times\text{v}\times\text{x}}{\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=1-2\text{v}^2+\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=1-2\text{v}^2$
$\Rightarrow\ \frac{\text{dv}}{1-2\text{v}^2}=\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{\text{dv}}{\text{v}^2-\frac{1}2}=-2\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \int\frac{\text{dv}}{\big(\frac{1}{\sqrt2}\big)^2-\text{v}^2}=2\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{\sqrt2}2\log\bigg(\frac{\frac{1}{\sqrt2}+\text{v}}{\frac{1}{\sqrt2}-\text{v}}\bigg)=2\log\text{x}+\log\text{C}$
$\Rightarrow\ \frac{1}{\sqrt2}\log\Bigg(\frac{\frac{1}{\sqrt2}+\frac{\text{y}}{\text{x}}}{\frac{1}{\sqrt2}-\frac{\text{y}}{\text{x}}}\Bigg)2\log\text{x}+\log\text{C}$
$\Rightarrow\ \frac{1}{\sqrt2}\log\Big(\frac{\text{x + y}\sqrt2}{\text{x}-\text{y}\sqrt2}\Big)2\log\text{x}+\log\text{C}$
$\Rightarrow\ \frac{1}{\sqrt2}\log\Big(\frac{\text{x + y}\sqrt2}{\text{x}-\text{y}\sqrt2}\Big)\log\text{x}^2+\log\text{C}$
$\Rightarrow\ \log\Big(\frac{\text{x + y}\sqrt2}{\text{x}-\text{y}\sqrt2}\Big)^{\frac{1}{\sqrt2}}=\log\text{Cx}^2$
$\Rightarrow\ \Big(\frac{\text{x + y}\sqrt2}{\text{x}-\text{y}\sqrt2}\Big)^{\frac{1}{\sqrt2}}=\text{Cx}^2$
$\Rightarrow\ \Big(\frac{\text{x + y}\sqrt2}{\text{x}-\text{y}\sqrt2}\Big)=\big(\text{Cx}^2\big)^{\sqrt2}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "4 Marks" from Differential Equations→Differential Equations — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Using the method of integration find the area bounded by the curve |x| + |y| = 1
[Hint: the required region is bounded by lines x + y = 1, x - y = 1, -x + y = 1 and -x - y = 11]
→
Show that $\text{f}(\text{x})=\frac{1}{1+\text{x}^2}$ is neither increasing nor decreasing on R.
→
A speaks truth in 60% of the cases, while B in 90% of the cases. In what percent of cases are they likely to contradict each other in stating the same fact? In the cases of contradiction do you think, the statement of B will carry more weight as he speaks truth in more number of cases than A?
→
Find the maximum slope of the curve y = −x3 + 3x2 + 2x − 27.
→
For the following matrices verify the associativity of multiplication i.e., (AB) C = A(BC):
$\text{A}=\begin{bmatrix}4&2&3\\1&1&2\\3&0&1\end{bmatrix},\text{B}=\begin{bmatrix}1&-1&1\\0&1&2\\2&-1&1\end{bmatrix}$ and $\text{C}=\begin{bmatrix}1&2&-1\\3&0&1\\0&0&1\end{bmatrix}$
→
If f'(x) = x + b, f'(1) = 5, f'(2) = 13, find f'(x).
→
Examine the differentiability of f, where f is defined by:
$\text{f(x)}=\begin{cases}1+\text{x},&\text{if x}\leq2\\5-\text{x},&\text{if x}>2\end{cases}$
at x = 2.
→
Prove that $2\tan^{-1}\bigg(\sqrt{\frac{\text{a}-\text{b}}{\text{a}+\text{b}}}\tan\frac{\theta}{2}\bigg)=\cos^{-1}\Big(\frac{\text{a}\cos\theta+b}{\text{a}+\text{b}\cos\theta}\Big)$
→
A swimming pool is to be drained for cleaning. If L represents the number of litres of water in the pool t seconds after the pool has been plugged off to drain and L = 200(10 - t)2. How fast is the water running out at the end of 5 seconds? What is the average rate at which the water flows out during the first 5 seconds?
→
If $\text{P}=\begin{bmatrix}\text{x}&0&0\\0&\text{y}&0\\0&0&\text{z}\end{bmatrix}$ and $\text{Q}=\begin{bmatrix}\text{a}&0&0\\0&\text{b}&0\\0&0&\text{c}\end{bmatrix},$ prove that $\text{PQ}=\begin{bmatrix}\text{xa}&0&0\\0&\text{y}\text{b}&0\\0&0&\text{zc}\end{bmatrix}=\text{QP}$
→