Question

y-x dy/dx = a(y^2 +dy/dx)

Answer 1

$\underline{\textsf{Question :}}$

$\textsf{Solve :}\:\mathsf{y-x\frac{dy}{dx}=a(y^{2}+\frac{dy}{dx})}$

$\underline{\textsf{Solution :}}$

$\textsf{Now,}\:\mathsf{y-x\frac{dy}{dx}=a(y^{2}+\frac{dy}{dx})}$

$\to \mathsf{y-x\frac{dy}{dx}=ay^{2}+a\frac{dy}{dx}}$

$\to \mathsf{x\frac{dy}{dx}+a\frac{dy}{dx}+ay^{2}-y=0}$

$\to \mathsf{(x+a)\frac{dy}{dx}+ay^{2}-y=0}$

$\to \mathsf{\frac{dy}{ay^{2}-y}+\frac{dx}{x+a}=0}$

$\to \mathsf{\frac{dy}{y^{2}-\frac{y}{a}}+\frac{a\:dx}{x+a}=0}$

$\to \mathsf{\frac{dy}{(y-\frac{1}{2a})^{2}-(\frac{1}{2a})^{2}}+\frac{a\:dx}{x+a}=0}$

$\to \mathsf{\frac{d(y-\frac{1}{2a})}{(y-\frac{1}{2a})^{2}-(\frac{1}{2a})^{2}}+\frac{a\:dx}{x+a}=0}$

$\textsf{On integration, we get}$

$\int \mathsf{\frac{d(y-\frac{1}{2a})}{(y-\frac{1}{2a})^{2}-(\frac{1}{2a})^{2}}+\int \frac{a\:dx}{x+a}=0}$

$\to \mathsf{\frac{1}{2*\frac{1}{2a}}log_{e}|\frac{y-\frac{1}{2a}-\frac{1}{2a}}{y-\frac{1}{2a}+\frac{1}{2a}}|+log_{e}|x+a|=C}$

$\textsf{where C is integral constant}$

$\to \mathsf{a\:log_{e}|\frac{1-\frac{1}{a}}{y}|+a\:log_{e}|x+a|=C}$

$\to \mathsf{log_{e}|\frac{ay-1}{ay}|+log_{e}|x+a|=\frac{C}{a}}$

$\to \mathsf{log_{e}|\frac{(ay-1)(x+a)}{ay}|=\frac{C}{a}}$

$\to \mathsf{\frac{(ay-1)(x+a)}{ay}=e^{\frac{C}{a}}}$

$\to \mathsf{\frac{(ay-1)(x+a)}{y}=a\:e^{\frac{C}{a}}}$

$\to \mathsf{\frac{(ay-1)(x+a)}{y}=k}$

where $\mathsf{k=a\:e^{\frac{C}{a}}}$

$\to \boxed{\mathsf{(ay-1)(x+a)=ky}}$ ,

$\textsf{which is the required solution.}$

Answer 2

Answer:$(a+x)(1-ay)=yc$

Step-by-step explanation:

We have,

$y-x\frac{dy}{dx}=a(y^2+\frac{dy}{dx})$

$y-x\frac{dy}{dx}=ay^2+a\frac{dy}{dx}$

$a\frac{dy}{dx}+x\frac{dy}{dx}=y-ay^2$

$(a+x)\frac{dy}{dx}=y(1-ay)$

$\frac{dy}{y(1-ay)}=\frac{dx}{a+x}$

$(\frac{1}{y}+\frac{a}{1-ay})dy=\frac{dx}{a+x}\;\;\;\;\text{(by P.F.)}$

$\frac{dy}{y}+\frac{a\;dy}{1-ay}=\frac{dx}{a+x}$

$\text{Integrating both sides with}\\\;\\\int{\frac{dy}{y}}+\int{\frac{a\;dy}{1-ay}}=\int{\frac{dx}{a+x}}\\\;\\\log{y}-\log{(1-ay)}=\log{(a+x)}+\log c\\\;\\\frac{y}{1-ay}=\frac{a+x}{c}$

$yc=(a+x)(1-ay)\\\;\\(a+x)(1-ay)=yc$

Note:-, Let,

$\frac{1}{y(1-ay)}=\frac{p}{y}+\frac{q}{1-ay}\\\;\\\frac{1}{y(1-ay)}=\frac{p-ayp+yq}{y(1-ay)}\\\;\\\frac{1}{y(1-ay)}=\frac{(q-pa)y+p}{y(1-ay}\\\;\\On,\;comparing\;\\\;\\p=1\\\;\\q-pa=0\\\;\\q-a=0\\\;\\q=a$