Differentials:Please to if my work is correct

I just want to see if my work is correct.

This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.





4.8 #1
y`` + y = 0

a.



e^xsolve with: e^x
e^x y`` + y = 0

a y"+b y' +c y=0

a y’’ +b y’ + c =0

a e^(y’’ x) + b e^(y’ x)

e^(y)(a cos(x) +b sin(x))

This is not a solution


b.Sin x
sin x y'' + y = 0


1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
y(x)+sin(x) = cos(x) y''(x)
Solution


c.Cos x
cos x y'' + y = 0

1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution






d.Sin x – cos x

-1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0


Solution

Quote:

---

Originally Posted by pikachumanson

I just want to see if my work is correct.

This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.





4.8 #1
y`` + y = 0

a.



e^xsolve with: e^x
e^x y`` + y = 0

a y"+b y' +c y=0

a y’’ +b y’ + c =0

a e^(y’’ x) + b e^(y’ x)

e^(y)(a cos(x) +b sin(x))

This is not a solution


b.Sin x
sin x y'' + y = 0


1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
y(x)+sin(x) = cos(x) y''(x)
Solution


c.Cos x
cos x y'' + y = 0

1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution






d.Sin x – cos x

-1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0


Solution

---

Sorry but I can't make any sense of this. Please post the questions exactly as they are written from wherever you got them from.

[QUOTE=pikachumanson;533340]I just want to see if my work is correct.

This problems comes in for parts. I am concerned about the last three. I think the processes should be longer but I don't see how I can make them longer.





4.8 #1
y`` + y = 0

Quote:

---

Is this the differential equation you want to solve?

[quote[ a.



e^xsolve with: e^x

---

What does that mean?

Quote:

---

e^x y`` + y = 0

---

Just put in front of y''? Why?

Quote:

---

a y"+b y' +c y=0

a y’’ +b y’ + c =0

a e^(y’’ x) + b e^(y’ x)

---

Now you have taken the exponential of each derivative. Again, why? And where did a, b, and c suddenly come from?

Quote:

---

e^(y)(a cos(x) +b sin(x))

This is not a solution


b.Sin x
sin x y'' + y = 0


1/2 i e^-i^x y``(x)- 1/2 i e^i^x y``(x)+y(x)=0
y(x)+sin(x) = cos(x) y''(x)
Solution


c.Cos x
cos x y'' + y = 0

1/2 e^(-i x) y''(x)+1/2 e^(i x) y''(x)+y(x) = 0cos(x) y''(x)+y(x) = 0Solution






d.Sin x – cos x

-1/2 e^(-i x) y''(x)-1/2 e^(i x) y''(x)+y(x)+1/2 i e^(-i x)-1/2 i e^(i x) = 0


Solution

---

I agree with mr. fantastic- what you wrote makes no sense at all!