Download App

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS4 Marks
Question
Solve the following differential equation:
$(\text{x}^2+3\text{xy}+\text{y}^2)\text{dx}-\text{x}^2\text{dy}=0$
✓

Answer

We have,
$(\text{x}^2+3\text{xy}+\text{y}^2)\text{dx}-\text{x}^2\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2+3\text{xy}+\text{y}^2}{\text{x}^2}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+3\text{vx}^2+\text{v}^2\text{x}^2}{\text{x}^2}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=1+3\text{v}+\text{v}^2-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=1+\text{v}^2+2\text{v}$
$\Rightarrow\ \frac{1}{1+\text{v}^2+2\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{1+\text{v}^2+2\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{(1+\text{v})^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ -\frac{1}{(1+\text{v})}=\log|\text{x}|+\text{C}$
$\Rightarrow\ \log|\text{x}|+\frac{1}{(1+\text{v})}=-\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\therefore\ \log|\text{x}|+\frac{\text{x}}{(\text{x + y})}=\text{C}_1$
where
C1 = -C
Hence, $\log|\text{x}|+\frac{\text{x}}{(\text{x + y})}=\text{C}_1$ is the required solution.

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

If $\text{x}-\text{e}^{\tan\text{x}}+\sqrt{\frac{\text{x}^2+1}{2}},$ find $\frac{\text{dy}}{\text{dx}}$
→
Differentiate the following functions with respect to x:
$\sin^{-1}\Big\{\frac{\sin\text{x}+\cos\text{x}}{\sqrt{2}}\Big\},-\frac{3\pi}{4}<\text{x}<\frac{\pi}{4}$
→
Using elementary transformation, find the inverse of each of the matrices, $\begin{bmatrix}2& 0&-1 \\5 & 1&0\\0&1&3 \end{bmatrix}$
→
Evaluate the following integrals as limit of sum:
$\int\limits^3_{1}(2\text{x}+3)\text{dx}$
→
Express the matrix $\text{A}=\begin{bmatrix}3 & -4 \\1 & -1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
→
Show that the four points A, B, C and D with the position vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ and $\vec{\text{d}}$ respectively are coplanar if and only if $3\vec{\text{a}}-2\vec{\text{b}}+\vec{\text{c}}-2\vec{\text{d}}=\vec0$.
→
Evaluate the following integrals as limit of sum:
$\int\limits^\text{b}_{\text{a}}\cos\text{x dx}$
→
If a1, a2, a3, ..., an is an arithmetic progression with common difference d, then evaluate the following expression.
$\tan\bigg[\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_1\text{a}_2}\Big)+\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_2\text{a}_3}\Big)+\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_3\text{a}_4}\Big)+\ .....\ +\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_{\text{n}-1}\text{a}_\text{n}}\Big)\bigg]$
→
Using properties of determinants, prove that
$\begin{vmatrix} \text{a}^{2} + \text{2a} & \text{2a + 1} & 1 \\ \text{2a + 1} & \text{a + 2} & 1 \\ 3 & 3 & 1 \end{vmatrix} = \text{(a - 1)}^{3}$
→
If $\text{A}=\begin{bmatrix}1&5\\7&12\end{bmatrix}$ and $\text{B}=\begin{bmatrix}9&1\\7&8\end{bmatrix},$ find a matrix C such that 3A + 5B + 2C is a null matrix.
→