# Can someone check my answers?

• Apr 14th 2010, 06:07 PM
deragon999
Hey people,

I would just like a few of my answers to be checked, if anyone wants to,

1) y'=e(-y)-1

I got ln(1-e(y))+x=c, but i think this is wrong(it doesn't match the given answer)

My working was:

dy/dx=1/e(y)-1
dy/dx=(1-e(y))/e(y)
int (1-e(y))/e(y) dy -int 1 dx=0
ln(1-e(y))+x=c

2) (dy/dx)tanx+tany=0

I got sinysinx=c

3) Find the general solution of:

d2y/dx2-5dy/dx+4y=2e(3x)

I got: Ae(4x)+Be(x)-e(3x)

4) d2y/dt2+4y=12cos(2t)

i) find Yc:

I got Yc=Acos(2t)+Bsin(2t)

ii) find the full general solution:

I got y=Acos(2t)+Bsin(2t)+3tsin(2t)

iii) given y'(0)=0 and y(0)=0 find the full solution

I got y=3tsin(2t)

5) evaluate int e(2x)sinx dx via trap rule, simpsons 1/3rd and gauss integration. With 4 strips for trapezoidal/simpsons.

I got:

Trap: 23.02829

Simp: 22.70988612

Gauss: 22.768

Any answer checks would be greatly appreciated, especially for the first question.
• Apr 14th 2010, 06:28 PM
Sudharaka
Quote:

Originally Posted by deragon999
Hey people,

I would just like a few of my answers to be checked, if anyone wants to,

1) y'=e(-y)-1

I got ln(1-e(y))+x=c, but i think this is wrong(it doesn't match the given answer)

My working was:

dy/dx=1/e(y)-1
dy/dx=(1-e(y))/e(y)
int (1-e(y))/e(y) dy -int 1 dx=0
ln(1-e(y))+x=c

2) (dy/dx)tanx+tany=0

I got sinysinx=c

3) Find the general solution of:

d2y/dx2-5dy/dx+4y=2e(3x)

I got: Ae(4x)+Be(x)-e(3x)

4) d2y/dt2+4y=12cos(2t)

i) find Yc:

I got Yc=Acos(2t)+Bsin(2t)

ii) find the full general solution:

I got y=Acos(2t)+Bsin(2t)+3tsin(2t)

iii) given y'(0)=0 and y(0)=0 find the full solution

I got y=3tsin(2t)

5) evaluate int e(2x)sinx dx via trap rule, simpsons 1/3rd and gauss integration. With 4 strips for trapezoidal/simpsons.

I got:

Trap: 23.02829

Simp: 22.70988612

Gauss: 22.768

Any answer checks would be greatly appreciated, especially for the first question.

Dear deragon999,

I suppose you mean $\displaystyle y'=e^{-y}-1$ in the first question. Is'nt??

If that is so,

$\displaystyle \frac{dy}{dx}=e^{-y}-1$

$\displaystyle \frac{dy}{dx}=\frac{1-e^{y}}{e^y}$

$\displaystyle \int{\frac{e^y}{1-e^y}}dy=\int{dx}$

$\displaystyle -\int{\frac{e^y}{e^y-1}}dy=\int{dx}$

$\displaystyle -\int{\frac{d}{dy}(e^{y}-1)\frac{d}{d(e^y-1)}ln(e^y-1)}dy=\int{dx}$

$\displaystyle -\int{d(ln(e^y-1))}=\int{dx}$

$\displaystyle -ln(e^y-1)=x+C$ ; Where C is an arbitary constant.

• Apr 14th 2010, 06:32 PM
deragon999
Sudharaka

Thanks, can you show me how you got

http://www.mathhelpforum.com/math-he...d2ca263c-1.gif

Thanks,
Simon
• Apr 14th 2010, 06:37 PM
Sudharaka
Quote:

Originally Posted by deragon999
Sudharaka

Thanks, can you show me how you got

http://www.mathhelpforum.com/math-he...d2ca263c-1.gif

Thanks,
Simon

Dear deragon999,

$\displaystyle \frac{dy}{dx}=e^{-y}-1=\frac{1}{e^y}-1=\frac{1-e^y}{e^y}$

• Apr 15th 2010, 06:33 PM
deragon999
Haha not relly, that bit i had already done in my working, the bit that you did different to me was taking -1 out of the dy integral

Thanks you for all your time spent anyway