# System of ODE without constant coefficients

I've got following system of ODE...

$$y_1'(x) = (3x-1)y_1(x)-(1-x)y_2(x)$$ $$y_2'(x)=-(x+2)y_1(x)+(x-2)y_2(x)$$

I'd like to have some tips on how to solve that. I've researched a lot on the Internet and on this site here but the message is always the same "It's hard to solve that but in this case, it's easy because you can use trick xy".

Well I don't see the trick here. Only thing I can think of, is to solve the first equation for $$y_2$$, so that...

$$y_2 = \frac{3x-1}{1-x}y_1(x) - \frac{1}{1-x}y_1'$$

and then differentiate so that I get $$y_2' = ...$$ and plug $$y_2, y_2'$$ into the second equation.

But that cant be a good idea, can it?

I'm hoping for some helpful tips. Thanks in advance =)

• I haven't gone through any of the details, but have you considered using a Laplace Transform? Maybe you can transform this system of ODEs into a system of equations. Do you have an initial condition? – Matthew Pilling Nov 29 '20 at 1:35
• No, I'm trying to do it without Laplace. What do you mean by transforming it into a system of equations? No, I don't have an initial condition. – Quotenbanane Nov 29 '20 at 1:44

## 1 Answer

$$y_1'(x) = (3x-1)y_1(x)-(1-x)y_2(x)$$ $$y_2'(x)=-(x+2)y_1(x)+(x-2)y_2(x)$$ Sum both DE: $$(y_1+y_2)'(x)=(2x-3)y_1(x)+(2x-3)y_2(x)$$ This is a separable differential equation. $$u'=(2x-3)u$$ Where $$u=y_1+y_2$$. $$(\ln u)'=2x-3$$ $$\ln u=x^2-3x+C$$ $$y_1+y_2=C_1e^{x^2-3x}$$ $$y_1=-y_2+C_1e^{x^2-3x}$$ Plug this in the second DE and solve.