Question

Find the equation of the curve passing through the point (1, 1), given that the slope of the tangent to the curve to the any point is $\frac{y}{x}+1$

Answer 1

Let the curve be $F(x,y)$ and $\frac{dy}{dx}$ be the slope.

Hence, the equation is $\frac{dy}{dx}=\frac{y}{x}+1$

$⟹\frac{dy}{dx}-\frac{y}{x}=1$ is the first order linear equation, where $P\left(x\right)=-\frac{1}{x}$ and $Q\left(x\right)=1$

$\therefore \text{I.F.}=e^{-\int \frac{1}{x}}dx=e^{-\log x}$

$\therefore$ Solution is $y{e}^{-\log x}=\int 1.e^{-\log x}dx$

$=\int x^{-1}dx$

$=\log x+c$

$⟹y{x}^{-1}=\log x+c$ ........ $\left(i\right)$

To find $c$, substitute the values of $x$ and $y$ we get

$1\cdot 1=\log 1+c$

$⟹c=1$

Put in $\left(i\right)$, we get

$y{x}^{-1}=\log x+1$

$⟹y=x(\log x+1)$ is the required equation.