Question

Find all curves y=y(x) such that the tangent to the curve at any point passes through a given point (x1, y1).

Answer 1

@SithsAndGiggles

Answer 2

For any point on the curve \((x_0,y_0)\), the slope of the tangent line to the curve is \(y'(x_0)\). The equation of the tangent line then is
\[y-y_0=y'(x_0)(x-x_0)\]
You know that the tangent line is to pass through a given point \((x_1,y_1)\), and so
\[y_1-y_0=y'(x_0)(x_1-x_0)\]
Since \(y=y(x)\) means \(y_0=y(x_0)\), you get an ODE that is linear in \(y(x_0)\).
\[y_1-y(x_0)=y'(x_0)(x_1-x_0)\]
or equivalently,
\[y'(x_0)+\frac{1}{x_1-x_0}y(x_0)=\frac{y_1}{x_1-x_0}\]

Answer 3

So I got dy/dx=(y1-y)/(x1-x), is this right? Do I solve for y from here?

Answer 4

Yes if you replace \(y(x_0)\) with \(y\) you get the same ODE.

Answer 5

So if dy/dx=(y1-y)/(x1-x), then how do you solve for y from here?
\[\frac{ dy }{ y _{1}-y }=\frac{ dx }{ x ^{1}-x }\]?

Answer 6

You can certainly treat this as a separable equation, so integrating both sides gives
\[\int\frac{dy}{y_1-y}=-\ln|y_1-y|=\ln\frac{1}{|y_1-y|}\]
and
\[\int\frac{dx}{x_1-x}=-\ln|x_1-x|+C=\ln\frac{1}{|y_1-y|}+C\]
Exponentiate both sides:
\[\exp\left(\ln\frac{1}{y_1-y}\right)=\exp\left(\ln\frac{1}{|y_1-y|}+C\right)\]
which gives
\[\frac{1}{y_1-y}=\frac{C}{x_1-x}\]

Answer 7

Alternatively, since the equation is linear in \(y(x_0)\), you can find the appropriate integrating factor.
\[y'(x_0)+\frac{1}{x_1-x_0}y(x_0)=\frac{y_1}{x_1-x_0}\]
suggests that the IF is
\[\mu(x_0)=\exp\left(\int\frac{dx_0}{x_1-x_0}\right)=\frac{1}{x_1-x_0}\]
This yields
\[\begin{align*}\frac{1}{x_1-x_0}y'(x_0)+\frac{1}{(x_1-x_0)^2}y(x_0)&=\frac{y_1}{(x_1-x_0)^2}\\\\
\frac{d}{dx_0}\left[\frac{1}{x_1-x_0}y(x_0)\right]&=\frac{y_1}{(x_1-x_0)^2}\\\\
\frac{1}{x_1-x_0}y(x_0)&=\int\frac{y_1}{(x_1-x_0)^2}\,dx_0\\\\
\frac{1}{x_1-x_0}y(x_0)&=\frac{y_1}{x_1-x_0}+C
\end{align*}\]

Answer 8

So \[y _{1}-y=\frac{ x _{1}-x }{ c }\]

Answer 9

Never mind. I got it! I got the right answer! Thank you!