Mathematics - Differential Equation Question with Solution | TestHub

Let the slope of the tangent at point $P(x,y)$ be $m=\frac{dy}{dx}$. The equation of the tangent line is $Y-y=m(X-x)$, which can be written as $mX-Y+(y-mx)=0$. The length of the perpendicular from the origin $(0,0)$ to this tangent line is given by $L=\frac{|m(0)-(0)+(y-mx)|}{\sqrt{m^2+(-1{)}^2}}=\frac{|y-mx|}{\sqrt{1+m^2}}$. According to the problem statement, this length is equal to the abscissa of $P$, i.e., $|x|$. So, $\frac{|y-mx|}{\sqrt{1+m^2}}=|x|$. Squaring both sides, we get $(y-mx{)}^2=x^2(1+m^2)$. Substituting $m=\frac{dy}{dx}$: $(y-x\frac{dy}{dx}{)}^2=x^2(1+(\frac{dy}{dx}{)}^2)$.

This is a homogeneous differential equation. Let $y=vx$, so $\frac{dy}{dx}=v+x\frac{dv}{dx}$. Substituting these into the equation:$(vx-x(v+x\frac{dv}{dx}){)}^2=x^2(1+(v+x\frac{dv}{dx}{)}^2)$$(-x^2\frac{dv}{dx}{)}^2=x^2(1+v^2+x^2(\frac{dv}{dx}{)}^2+2vx\frac{dv}{dx})$$x^4(\frac{dv}{dx}{)}^2=x^2+x^2{v}^2+x^4(\frac{dv}{dx}{)}^2+2v{x}^3\frac{dv}{dx}$. Dividing by$x^2$ (assuming $x\ne 0$): $x^2(\frac{dv}{dx}{)}^2=1+v^2+x^2(\frac{dv}{dx}{)}^2+2vx\frac{dv}{dx}$. This simplifies to$0=1+v^2+2vx\frac{dv}{dx}$. Rearranging, $2vx\frac{dv}{dx}=-(1+v^2)$. This is a variable separable equation: $\frac{2v}{1+v^2}dv=-\frac{1}{x}dx$. Integrating both sides: $\int \frac{2v}{1+v^2}dv=-\int \frac{1}{x}dx$.$\ln (1+v^2)=-\ln |x|+C_1=\ln (\frac{1}{|x|})+C_1$.$1+v^2=e^{C_1}\frac{1}{|x|}=\frac{A}{|x|}$ where $A=e^{C_1}$. Substituting $v=\frac{y}{x}$: $1+(\frac{y}{x}{)}^2=\frac{A}{|x|}$.$\frac{x^2+y^2}{x^2}=\frac{A}{|x|}$. Assuming $x>0$, $x^2+y^2=Ax$. The curve passes through $(1,1)$. So, $1^2+1^2=A(1)\Rightarrow A=2$. Thus, the equation of the curve is $x^2+y^2=2x$ (or $(x-1{)}^2+y^2=1$, a circle).