1. ## Can someone check my answers?

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.

2. 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 $y'=e^{-y}-1$ in the first question. Is'nt??

If that is so,

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

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

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

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

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

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

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

3. Sudharaka

Thanks, can you show me how you got

Thanks,
Simon

4. Originally Posted by deragon999
Sudharaka

Thanks, can you show me how you got

Thanks,
Simon
Dear deragon999,

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