The solution of x dx + y dy = x^2 y dy - x y^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the differential equation: $x dx + y dy = x^2 y dy - x y^2 dx$ Rearranging the terms to group $x$ and $y$ variables: $x dx + x y^2 dx = x^2

Vedclass · Accepted Answer

$x^2 - 1 = C(1 + y^2)$

Vedclass · Answer

(A) Given the differential equation: $x dx + y dy = x^2 y dy - x y^2 dx$ Rearranging the terms to group $x$ and $y$ variables: $x dx + x y^2 dx = x^2 y dy - y dy$ $x(1 + y^2) dx = y(x^2 - 1) dy$ Separating the variables: $\frac{x}{x^2 - 1} dx = \frac{y}{1 + y^2} dy$ Multiplying both sides by $2$: $\frac{2x}{x^2 - 1} dx = \frac{2y}{1 + y^2} dy$ Integrating both sides: $\int \frac{2x}{x^2 - 1} dx = \int \frac{2y}{1 + y^2} dy$ $\ln|x^2 - 1| = \ln|1 + y^2| + \ln C$ Using the property $\ln a + \ln b = \ln(ab)$: $\ln|x^2 - 1| = \ln|C(1 + y^2)|$ Removing the logarithms: $x^2 - 1 = C(1 + y^2)$

List-$I$	List-$II$
$A$. Figure of merit $(k)$	$P$. in series
$B$. Current sensitivity $(S_I)$	$Q$. to get power loss minimized
$C$. Deflection is made half by using resistance	$R$. in shunt/parallel
$D$. High resistance box connected	$S$. current per unit deflection
	$T$. deflection per unit current

The solution of $x dx + y dy = x^2 y dy - x y^2 dx$ is

Similar Questions

The locus of the midpoints of the chords of the parabola $y^2 = 4ax$ which pass through the origin is:

The power dissipated in the circuit shown in the figure is $30 \, W$. The value of $R$ is ............. $\Omega$.

An organ pipe $P_1$ closed at one end is vibrating in its first overtone. Another pipe $P_2$ open at both ends is vibrating in its third overtone. They are in resonance with a given tuning fork. The ratio of the length of $P_1$ to that of $P_2$ is:

$A$ car weighs $1800\,kg$. The distance between its front and back axles is $1.8\,m$. Its centre of gravity is $1.05\,m$ behind the front axle. Force exerted by the level ground on each front wheel is ........ $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module