# Thread: [SOLVED] Stumped on this DE

1. ## [SOLVED] Stumped on this DE

Solve.

$\int 7x cos(x) = \int (2y + e^{3y}) y'$, $y(0) = 0$

I'm pretty sure my work is right, but as usual Webassign does not care for my answer.
$
-7 cos(x) - 7x sin(x) + C = y^2 + \frac{e^{3y}}{3}$

*Now, webassign is having me give the answer based on what and how they provide HALF of the answer. So, here is what WA has on the left:

$e^{3y} + 3y^2$ = __ my answer________

So, before I solved for C, I went ahead and multiplied through by the 3 so that I could just work it through in the form it has to be in anyway:

$3y^2 + e^{3y} = 21 (-xsin(x) - cos(x)) + 3C$

yadda yadda $C = \frac{22}{3}$

So, the right side answer that I am getting is:

$-21x sin(x) - 21 cos(x) + \frac{22}{3}$

Anyone? Anyone? Thanks!

2. $\int 7x\cos{x} \, dx = 7(\cos{x} + x\sin{x}) + C$

$\int 2y + e^{3y} \, dy = y^2 + \frac{1}{3}e^{3y} + C$

$7(\cos{x} + x\sin{x}) + C = y^2 + \frac{1}{3}e^{3y}$

$21(\cos{x} + x\sin{x}) + C = 3y^2 + e^{3y}$

$y(0) = 0$

$21(1) + C = 1$

$C = -20$

$21(\cos{x} + x\sin{x}) - 20 = 3y^2 + e^{3y}$

3. Originally Posted by skeeter
$\int 7x\cos{x} \, dx = 7(\cos{x} + x\sin{x}) + C$

$\int 2y + e^{3y} \, dy = y^2 + \frac{1}{3}e^{3y} + C$

$7(\cos{x} + x\sin{x}) + C = y^2 + \frac{1}{3}e^{3y}$

$21(\cos{x} + x\sin{x}) + C = 3y^2 + e^{3y}$

$y(0) = 0$

$21(1) + C = 1$

$C = -20$

$21(\cos{x} + x\sin{x}) - 20 = 3y^2 + e^{3y}$

So you don't make C, 3C when you multiply through with that 3?

4. Originally Posted by mollymcf2009
I'm pretty sure my work is right, but as usual Webassign does not care for my answer.
$
-7 cos(x) - 7x sin(x) + C = y^2 + \frac{e^{3y}}{3}$
Not quite.

$\int7x\cos x\,dx$

$=\int7x(\sin x)'\,dx$

$=7x\sin x-7\int\sin x\,dx$

$=7x\sin x+7\cos x+C_0$

$=7(x\sin x+\cos x)+C_0.$

$3y^2 + e^{3y} = 21 (-xsin(x) - cos(x)) + 3C$
So we should have

$7(x\sin x+\cos x)+C_0=y^2+\frac{e^{3y}}3$

Multiplying by 3, you should get

$21(x\sin x+\cos x)+C=3y^2+e^{3y}$ (where $C=3C_0$)

which gives $C=-20.$

So, the right side answer that I am getting is:

$-21x sin(x) - 21 cos(x) + \frac{22}{3}$
Which, after fixing the mistakes, should be

$21(x\sin x+\cos x)-20.$

Edit:
So you don't make C, 3C when you multiply through with that 3?
Technically, yes. Skeeter was being a little informal with his notation, but his method is correct.

5. Originally Posted by mollymcf2009
So you don't make C, 3C when you multiply through with that 3?
no ... 3 times a Constant is just another Constant.

6. Originally Posted by mollymcf2009
So you don't make C, 3C when you multiply through with that 3?
C is an unknown constant so 3C is just another unknown constant.
Skeeter just used the same symbol to represent the two different constants.

7. Originally Posted by Reckoner
... Skeeter was being a little informal with his notation, but his method is correct.
I'll wear my tux next time ...

8. Originally Posted by skeeter
I'll wear my tux next time ...

9. Originally Posted by skeeter
I'll wear my tux next time ...
Y'all crack me up!