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Mathematics Question on General and Particular Solutions of a Differential Equation

The equation of the curve passing through the point (a,−1a)\left(a, -\frac{1}{a}\right) and satisfying the differential equation y−xdydx=a(y2+dydx)y-x \frac{dy}{dx}=a\left(y^{2}+\frac{dy}{dx}\right) is

A

(x+a)(1+ay)=−4a2y(x+ a)(1+a y)=-4 a^{2} y

B

(x+a)(1−ay)=4a2y(x+ a)(1-a y)=4 a^{2} y

C

(x+a)(1−ay)=−4a2y(x+ a)(1-a y)=-4 a^{2} y

D

None of these

Answer

(x+a)(1−ay)=−4a2y(x+ a)(1-a y)=-4 a^{2} y

Explanation

Solution

We have y−xdydx=a(y2+dydx)y-x \frac{d y}{d x}=a\left(y^{2}+\frac{d y}{d x}\right)
⇒ydx−xdy=ay2dx+ady\Rightarrow y d x-x d y=a y^{2} d x+ a d y
⇒y(1−ay)dx=(x+a)dy\Rightarrow y(1-a y) d x=(x +a) d y
⇒dxx+a−dyy(1−ay)=0\Rightarrow \frac{d x}{x+a}-\frac{d y}{y(1-a y)}=0
Integrating, we get
log⁡(x+a)−log⁡y+log⁡(1−ay)=log⁡C\log (x +a)-\log y+\log (1-a y)=\log C
or log⁡(a+x)(1−ay)y=log⁡C\log \frac{(a+ x)(1-a y)}{y}=\log C
i.e.(x+a)(1−ay)=Cy(x +a)(1-a y)=C y
Since the curve passes through (a,−1a)\left(a,-\frac{1}{a}\right)
∴2a×(1+1)=−Ca\therefore 2 a \times(1+1)=-\frac{C}{a}
i.e C=−4a2C=-4 a^{2}
So, (x+a)(1−ay)=−4a2y(x+ a)(1-a y)=-4 a^{2} y