# Thread: 5dy/dx + 2x = 3.

1. ## 5dy/dx + 2x = 3.

The problem is $\displaystyle 5\frac{dy}{dx}$ +2x = 3

by re-arranging I get the following

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

but need help with finding the general solution

i integrated an came up with.

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

p.s. what is the source code for integral sign?

2. \int is $\displaystyle \int.$

You should write your rearrangement as

$\displaystyle dy=\left(\dfrac{3}{5}-\dfrac{2x}{5}\right)dx,$ and then integrate. I'm not sure I buy your integration. Try again.

3. okey i have tried again...

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

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

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

can you show me where im goin wrong please. many thanks

4. Where did the 7 come from?

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

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

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

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

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

can you show me where im goin wrong please. many thanks

6. need to go back over integration. thanks guys

7. You're welcome for my contribution.