Solve the following differential equation: y^2dx+(x^2-xy+y^2)dy=0 - Vidyadip

Question

Solve the following differential equation:
$\text{y}^2\text{dx}+(\text{x}^2-\text{xy}+\text{y}^2)\text{dy}=0$

✓

Answer

We have,
$\text{y}^2\text{dx}+(\text{x}^2-\text{xy}+\text{y}^2)\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-\text{y}^2}{\text{x}^2-\text{xy}+\text{y}^2}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^2\text{x}^2}{\text{x}^2-\text{vx}^2+\text{v}^2\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^2}{1-\text{v}+\text{v}^2}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}^2}{1-\text{v}+\text{v}^2}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{-\text{v}-\text{v}^3}{1-\text{v}+\text{v}^2}$
$\Rightarrow\ \frac{1-\text{v}+\text{v}^2}{\text{v}+\text{v}^3}\text{dv}=-\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \frac{1+\text{v}^2-\text{v}}{\text{v}(1+\text{v}^2)}\text{dv}=-\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1+\text{v}^2-\text{v}}{\text{v}(1+\text{v}^2)}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1+\text{v}^2}{\text{v}(1+\text{v}^2)}\text{dv}-\int\frac{1}{\text{v}(1+\text{v}^2)}\text{dv}=-\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{1}{\text{v}}\text{dv}-\int\frac{1}{1+\text{v}^2}\text{dv}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log|\text{v}|-\tan^{-1}|\text{v}|=-\log|\text{x}|+\log\text{C}$
$\Rightarrow\log\big|\frac{\text{vx}}{\text{C}}\big|=\tan^{-1}\text{v}$
$\Rightarrow\ \big|\frac{\text{vx}}{\text{C}}\big|=\text{e}^{\tan^{-1}\text{v}}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ |\text{y}|=\text{Ce}^{\tan^{-1}\text{v}}$
Hence, $|\text{y}|=\text{Ce}^{\tan^{-1}\text{v}}$ 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 "5 Marks Questions" from Differential Equations→Differential Equations — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Find the values of a so that the function
$\text{f}\text{(x)}=\begin{cases}\text{ax}+5, &\text{if}\text{ x}\leq2\\\text{x}-1, &\text{if}\text{ x}>2\end{cases}$ is continuous at x = 2.

Discuss the commutativity and associativity of binary operation $^{‘*’}$ defined on A = Q – {1} by the rule $\text{a} ^{*} \text{b = a – b + ab}$ for all a, b $\in$ A. Also find the identity element of $^{*}$ in A and hence find the invertible elements of A.

Find the vector equation of the plane passing through the points P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3).

If O is a point in space, ABC is a triangle and D, E, F are the mid-points of the sides BC, CA and AB respectively of the triangle, prove that $\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}=\overrightarrow{\text{OD}}+\overrightarrow{\text{OE}}+\overrightarrow{\text{OF}}$.

Show that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero.

Integrate the function: $\frac{\sqrt{\tan x}}{\sin x \cos x}$

A chemical company produces two compounds, $A$ and $B.$ The following table gives the units of ingredients, $C$ and $D$ per kg of compounds $A$ and $B$ as well as minimum requirements of $C$ and $D$ and costs per kg of $A$ and $B.$ Find the quantities of $A$ and $B$ which would give a supply of $C$ and $D$ at a minimum cost.

	Compound		Minimum requirement
	$A$	$B$
Ingredient $C$	$1$	$2$	$80$
Ingredient $D$	$3$	$1$	$75$
Coist $($in $Rs.)$ per Kg	$4$	$6$

By using properties of determinants, show that:
$\begin{vmatrix}1+a^2-b^2&2ab&-2b\\2ab&1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2\end{vmatrix}=(1+a^2+b^2)^3$

Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Using mean value theorem, prove that there is a point on the curve $y = 2x^2 - 5x + 3$ between the points $A(1, 0)$ and $B(2, 1),$ where tangent is parallel to the chord AB. Also, find that point.