Ellen Chang

2022-07-13

What is the standard method for finding solutions of differential equations such as this one? (if there is any)

$x{y}^{\prime}={y}^{2}-(2x+1)y+{x}^{2}+2x$

where $y=ax+b$ is a particular solution.

Do I substitute $y$ with $ax+b+u(x)$ and then search for a solution or am I not noticing something and there's quicker way?

$x{y}^{\prime}={y}^{2}-(2x+1)y+{x}^{2}+2x$

where $y=ax+b$ is a particular solution.

Do I substitute $y$ with $ax+b+u(x)$ and then search for a solution or am I not noticing something and there's quicker way?

bap1287dg

Beginner2022-07-14Added 13 answers

I think that your hint is "there exists $a,b$ such thaht $y(x)=ax+b$ is solution". We find easily that $y(x)=x$ and $y(x)=x+1$ are solutions.

Hint: Note that your equation is

$x{y}^{\mathrm{\prime}}(x)-x=(y(x)-x)(y(x)-x-1)$

Now put $y(x)=x+z(x)$.

Hint: Note that your equation is

$x{y}^{\mathrm{\prime}}(x)-x=(y(x)-x)(y(x)-x-1)$

Now put $y(x)=x+z(x)$.

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I Completed the square on the bottom but what do you do now?

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My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$