Solving First Order Differential Equations - A Beginner's Guide

[suspenc3]

I am new to to this topic, hints?

$$y^1=x+5y$$

The only examples are in the form dy/dx+p[x(y)]=Q[x]

 

[StatusX]

Try some simple forms for the solution. y=ax+b should work.

If you want to be less reliant on luck, you could try looking at the asymptotic behavior of the DE. It seems to resemble the equation for exponential growth, in which case the y term would soon dominate over the x term on the RHS, and the solution would approach true exponential growth more and more closely. This suggests trying a solution of the form y=f(x) e^5x. Plugging this in and deriving a DE for f(x), you get something you can solve easily, but it turns out the exponential cancels, and you're left with something of the above form.

 

[HallsofIvy]

Do you mean the differential equation y'= x+ 5y?

You say "The only examples are in the form dy/dx+p[x(y)]=Q[x]". Surely you must mean dy/dx+ p(y(x))= Q(x). This is exactly of that form:
dy/dx- 5y= x. p(y)= 5y and Q(x)= x. StatusX's suggestion of trying y= f(x)e^5x is excellent but you could do basically the same thing by multiplying the entire equation by e^5x (an "integrating factor"):
$$e^{5x}\frac{dy}{dx}+ 5e^{5x}y= xe^{5x}$$
because
$$\frac{d(e^{5x}y)}{dx}= e^{5x}\frac{dy}{dx}+ 5e^{5x}y$$
by the product rule. The problem reduces to an integration by parts.

 

[StatusX]

Right, sorry. I've been doing DEs lately where there's no obvious solution, and these are the methods I've used. I forgot that equations of this form have a general method of solution, namely http://en.wikipedia.org/wiki/Integrating_factor" . Thanks Halls.

 

[suspenc3]

soo..$$\int \frac{d}{dx}(e^5^xy) - \int 5e^5^x = 5e^5^xy$$?