I am new to Differential Equations and am looking for some help regarding the following:

Find a differential equation the solutions to which describe the following curve: the sum of the tangential slope at point (x,y) on the curve and the slope between point (x,y) and (3,2) is exactly 1. Find the equation of the curve that passes through (0,0), while matching the criterion above.

I understand the theory, and I built the following solution: $y'=1-\frac{(y-2)}{x-3}$.

An online differential equations calculator gave the solution $y=\frac{x^2-2x+c_1}{2(x-3)}$.

As the curve has to pass through $(0,0)$, $c_1$ comes out 0, which leaves me with $y=\frac{x^2-2x}{2(x-3)}$.

While I think I have the right answer, what is bothering me is that I was not able to get from $y'$ to $y$ manually. It is my first week dealing with this topic, and I would like to gain this insight.

Many thanks!


On this one you can use an integration factor.

We have $y'+a(x)y=b(x)$, where $a(x)=\frac1{x-3}$ and $b(x)=1+\frac2{x-3}$.

Use $e^{\int a(x)\operatorname dx}=x-3$.

Multiply through and rewrite, using the product and chain rules.

Get $\frac{\operatorname d}{\operatorname dx}((x-3)y)=x-3+2=x-1$.

Integrate: $\int\frac{\operatorname d}{\operatorname dx}((x-3)y)=\int( x-1)$.

Then by FTC, we get $(x-3)y+C=\frac{x^2}2-x$.

So $y=\frac{x^2-2x}{2(x-3)}+C$.


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