# Thread: general solution to DE

1. ## general solution to DE

dy/dx +2y/x = 4x

find general solution

dy/dx= 4x + x/2y

2y dy = 4x + x dx

2y dy = 5x dx

integrate

y^2 = 2.5 x^2 + C

2. Originally Posted by harveyo
dy/dx +2y/x = 4x

find general solution

dy/dx= 4x + x/2y

2y dy = 4x + x dx

2y dy = 5x dx

integrate

y^2 = 2.5 x^2 + C
What you have done is not correct.

Your DE is not separable, you need to use the integrating factor method.

Do you know this method?

Read this http://www.mathhelpforum.com/math-he...equations.html

3. y' + 1/x 2y = 4x

so I(x) = e intergral 1/x dx = e ln x = x

now

x y' + 2y = 4x^2

4. Originally Posted by harveyo
y' + 1/x 2y = 4x

so I(x) = e intergral 1/x dx = e ln x = x

now

x y' + 2y = 4x^2
make it $y' + \frac{2}{x} ~y = 4x$

now find $I(x) = e^{\int \frac{2}{x}~dx}$

5. = e ln x^2 = x^2 ???

now

x^2 dy/dx + 2/x ^2 y = 4x^3

(x^2 y )' = 4x^3

y = x^2

6. Originally Posted by harveyo
= e ln x^2 = x^2 ???

now
Yep, great work, now multipling $x^2$ through the whole equation you get

$x^2 \times (y' + \frac{2}{x} ~y = 4x) = y'x^2+2xy= 4x^3$

and

$y'x^2+2xy= 4x^3$

by the product rule

$x^2y= \int 4x^3~dx$

Now can you take it from here?