Differential equations assignment T5

[mathi85]

Hi!

I would like to ask anyone with some spare time to check my assignment questions. Last time I was asked to post one task at a time so I will.
Thank you in advance for your time.

Task 5:

Find the particular solution of the following differential equations:
a) 12(d²y/dx²)-3y=0
given that: x=0, y=3 and (dy/dx)=0.5

b) (d²y/dx²)+2(dy/dx)+2y=10e^x
given that: x=0, y=0 and (dy/dx)=1

Solution:
a)
12(d²y/dx²)-3y=0 /:12
(d²y/dx²)-(1/4)y=0

m²=n²
m=+/-n

∴y=Ae^nx+Be^-nx
or y=Acosh(nx)+Bsinh(nx)

m²=1/4
∴m=+/-√(1/4)=+/-(1/2)

y=Acosh[(1/2)x]+Bsinh[(1/2)x]

3=Acosh[0]+Bsinh[0]
3=A(e⁰+e⁰/2)+B(e⁰-e⁰/2)
A=3

dy/dx=Asinh(0)+Bcosh(0)
B=1/2

Solution
b)
d²y/dx²+2(dy/dx)+2y=10e^x

m²+2m+2=0
∴m=-1+/-j

CF:
u=e^-x{Acos(x)+Bsin(x)}

PI:
v=pe^x
v'=pe^x
v''=pe^x

pe^x+2pe^x+2pe^x=10e^x
5pe^x=10e^x /:5e^x
p=10e^x/5e^x=2

v=2e^x

GS:
y=e^-x{Acos(x)+Bsin(x)}+2e^x [1]

dy/dx=e^-x{-Asin(x)+Bcos(x)}-e^-x{Acos(x)+Bsin(x)}+2e^x [2]

Sub into [1]
0=e⁰{Acos(0)+Bsin(0)}+2e⁰
A=-2

Sub into [2]
1=e⁰{2sin(0)+Bcos(0)}-e⁰{2cos(0)+Bsin(0)}+2e⁰
1=B-2+2
B=1

PS:
y=e^-x{-2cos(x)+sin(x)}+2e^x

 

[HallsofIvy]

Hi!

I would like to ask anyone with some spare time to check my assignment questions. Last time I was asked to post one task at a time so I will.
Thank you in advance for your time.

Task 5:

Find the particular solution of the following differential equations:
a) 12(d²y/dx²)-3y=0
given that: x=0, y=3 and (dy/dx)=0.5

b) (d²y/dx²)+2(dy/dx)+2y=10e^x
given that: x=0, y=0 and (dy/dx)=1

Solution:
a)
12(d²y/dx²)-3y=0 /:12
(d²y/dx²)-(1/4)y=0

m²=n²
m=+/-n

∴y=Ae^nx+Be^-nx
or y=Acosh(nx)+Bsinh(nx)

m²=1/4
∴m=+/-√(1/4)=+/-(1/2)

y=Acosh[(1/2)x]+Bsinh[(1/2)x]

3=Acosh[0]+Bsinh[0]
3=A(e⁰+e⁰/2)+B(e⁰-e⁰/2)
A=3

This is correct but I don't see why you would change to exponential form. If you are using hyperbolic functions you should know that a good reason for doing that is that cosh(0)= 1 and sinh(0)= 0. 3= Acosh(0)+ Bsinh(0) is immediately "3= A".

dy/dx=Asinh(0)+Bcosh(0)
B=1/2

Yes, dy/dx(0)= B= 0.5
Now, how about actually writing the solution to the problem?

Solution
b)
d²y/dx²+2(dy/dx)+2y=10e^x

m²+2m+2=0
∴m=-1+/-j

CF:
u=e^-x{Acos(x)+Bsin(x)}

PI:
v=pe^x
v'=pe^x
v''=pe^x

pe^x+2pe^x+2pe^x=10e^x
5pe^x=10e^x /:5e^x
p=10e^x/5e^x=2

v=2e^x

GS:
y=e^-x{Acos(x)+Bsin(x)}+2e^x [1]

dy/dx=e^-x{-Asin(x)+Bcos(x)}-e^-x{Acos(x)+Bsin(x)}+2e^x [2]

Sub into [1]
0=e⁰{Acos(0)+Bsin(0)}+2e⁰
A=-2

Sub into [2]
1=e⁰{2sin(0)+Bcos(0)}-e⁰{2cos(0)+Bsin(0)}+2e⁰

You've lost a sign here: the second term should be -e⁰(-2cos(0)+ Bsin(0))

1=B-2+2

So this should be 1= B+ 2+ 2

B=1

PS:
y=e^-x{-2cos(x)+sin(x)}+2e^x

 

[mathi85]

Thank you for the reply!

This is correct but I don't see why you would change to exponential form. If you are using hyperbolic functions you should know that a good reason for doing that is that cosh(0)= 1 and sinh(0)= 0. 3= Acosh(0)+ Bsinh(0) is immediately "3= A".

I'm actually not sure why I did it.

Yes, dy/dx(0)= B= 0.5
Now, how about actually writing the solution to the problem?

y=3cosh(x/2)+0.5sinh(x/2)

You've lost a sign here: the second term should be -e⁰(-2cos(0)+ Bsin(0))


So this should be 1= B+ 2+ 2

B=-3

∴y=e^-x{-2cos(x)-3sin(x)}+2e^x