# 5dy/dx + 2x = 3.

• Apr 6th 2011, 09:16 AM
decoy808
5dy/dx + 2x = 3.
The problem is $5\frac{dy}{dx}$ +2x = 3

by re-arranging I get the following

y= $\frac{3}{5} -\frac{2x}{5}$ dx

but need help with finding the general solution

i integrated an came up with.

y = $\frac{3^2}{6} - \frac{2x^2}{6}$ +C

p.s. what is the source code for integral sign?
• Apr 6th 2011, 09:19 AM
Ackbeet
\int is $\int.$

You should write your rearrangement as

$dy=\left(\dfrac{3}{5}-\dfrac{2x}{5}\right)dx,$ and then integrate. I'm not sure I buy your integration. Try again.
• Apr 6th 2011, 09:46 AM
decoy808
okey i have tried again...

$dy=\left(\dfrac{3x}{5}-\dfrac{?}{?}\right)dx$,

but cannot see where this 2nd part comes from.
$dy=\left(\dfrac{3x}{5}-\dfrac{2x^2}{7}\right)dx$
is different from the solution in the book. The solution in the book is

$dy=\left(\dfrac{3x}{5}-\dfrac{x^2}{5}\right)dx$

can you show me where im goin wrong please. many thanks :)
• Apr 6th 2011, 09:50 AM
Ackbeet
Where did the 7 come from?
• Apr 6th 2011, 10:14 AM
HallsofIvy
Quote:

Originally Posted by decoy808
okey i have tried again...

$dy=\left(\dfrac{3x}{5}-\dfrac{?}{?}\right)dx$,

Why the questions marks? You had before
$dy= \left(\dfrac{3}{5}- \dfrac{2x}{5}\right)dx$
Now integrate both sides: the integral of a constant is that constant times x and the integral of x is $(1/2)x^2$

Quote:

but cannot see where this 2nd part comes from.
$dy=\left(\dfrac{3x}{5}-\dfrac{2x^2}{7}\right)dx$
$\dfrac{3x}{5}$ is correct. The integral of $\frac{2}{5}x$ is $\left(\frac{2}{5}\right)\left(\frac{1}{2}x^2\right )= \frac{1}{5}x^2$
Perhaps you misread the $\frac{2}{5}\cdot\frac{1}{2}$ as $\frac{2}{5+ 2}$.

Quote:

is different from the solution in the book. The solution in the book is

$dy=\left(\dfrac{3x}{5}-\dfrac{x^2}{5}\right)dx$

can you show me where im goin wrong please. many thanks :)
• Apr 6th 2011, 10:20 AM
decoy808
need to go back over integration. thanks guys
• Apr 6th 2011, 12:15 PM
Ackbeet
You're welcome for my contribution.