Characteristic equation & imposing conditions

[c4l3b]

ODE

k^2 + 3k - 10 = 0
Roots: k1= 2 ; k2 = -5 >>>> How can I be sure those roots are suitable for equation because it could be vice versa?
y = Ae^3x + Be^-2x

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Impose conditions for particular solution;

y(0) = 5 gives 5 = A
y' (0) = gives 12 = -A + B

Whats the best way to calculate the conditions? perhaps by identities?

 

[galactus]

What's the ODE?. It would be best if you posted the original problem statement.

\(\displaystyle y''+3y'-10y=0\)

What are the given initial conditions?. y(0)=5 and y'(0)=?.

You are correct about the auxiliary equation \(\displaystyle m^{2}+3m-10=0\)

\(\displaystyle m=2, \;\ m=-5\)

\(\displaystyle y=C_{1}e^{2x}+C_{2}e^{-5x}\)

I will make up an I.C.

Let y(0)=1 and y'(0)=0

\(\displaystyle 1=C_{1}e^{2(0)}+C_{2}e^{-5(0)}\Rightarrow 1=C_{1}+C_{2}\)

\(\displaystyle C_{2}=1-C_{1}\)

y':

\(\displaystyle 0=2C_{1}e^{2(0)}+(1-C_{1})e^{-5(0)}\Rightarrow 0=7C_{1}-5\)

\(\displaystyle C_{1}=\frac{5}{7}, \;\ C_{2}=\frac{2}{7}\)

\(\displaystyle y=\frac{5}{7}e^{2x}+\frac{2}{7}e^{-5x}\)

It does not matter if you switch them to \(\displaystyle y=C_{1}e^{-5x}+C_{2}e^{2x}\)

It will work out and we solve using the initial conditions.

 

[c4l3b]

To find particular solution
y'' + 2y' + y = x^2 + 2

conditions y(0) = 3 and y(0) = 1

Since this a in-homogeneous equation do I just solve the RHS seperately x^2 and 2?
Would I need to express finding the particular function?

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y + y' = 3

x > = 0

conditions y(o) = 1


y = A cos x + B sin x

y' = -A sin x + B cos x and y'' = -A cos x - B sin x

I know this is the solution but its those damn conditions again, I mean if zero was replace with pi what would happen?

Advice please.

 

[galactus]

To find particular solution
y'' + 2y' + y = x^2 + 2

conditions y(0) = 3 and y(0) = 1

Since this a in-homogeneous equation do I just solve the RHS seperately x^2 and 2?
Would I need to express finding the particular function?

Are you familiar with LaPlace tranforms?. That is the method I prefer when doing these. But, you can use variations of parameters or some other method.

Is that y'(0)=1?. You have two y(0)'s. Are you required to use a particular method to solve?.
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y + y' = 3

x > = 0

conditions y(o) = 1


y = A cos x + B sin x

y' = -A sin x + B cos x and y'' = -A cos x - B sin x

I know this is the solution but its those damn conditions again, I mean if zero was replace with pi what would happen?

Advice please.

[/quote]

For this one, just separate variables:

\(\displaystyle \frac{dy}{dx}=3-y\)

\(\displaystyle \frac{dy}{3-y}=dx\)

\(\displaystyle \int\frac{1}{3-y}=\int dx\)

\(\displaystyle -ln(y-3)=x+c\)

\(\displaystyle y=C_{1}e^{-x}+3\)

Now, use y(0)=1 to find C1:

\(\displaystyle 1=C_{1}e^{-0}+3\)

\(\displaystyle 1=C_{1}+3\)

\(\displaystyle C_{1}=-2\)

\(\displaystyle y=-2e^{-x}+3\)

 

[c4l3b]

oh sorry about,

y'' + 2y' + y = x^2 + 2

conditions y(0) = 3 and y'(0) = 1

Btw I am not familiar with LaPlace tranforms.

 

[Deleted member 4993]

To find particular solution
y'' + 2y' + y = x^2 + 2

conditions y(0) = 3 and y(0) = 1

Since this a in-homogeneous equation do I just solve the RHS seperately x^2 and 2?
Would I need to express finding the particular function?

----------------------------------------------------------------------------------------

y + y' = 3

x > = 0

conditions y(o) = 1


y = A cos x + B sin x <<< Incorrect - Does it satisfy your given ODE (y" + 2y + y = x[sup:bup7lg6z]2[/sup:bup7lg6z] + 2)?

y' = -A sin x + B cos x and y'' = -A cos x - B sin x

I know this is the solution but its those damn conditions again, I mean if zero was replace with pi what would happen?

Advice please.

y'' + 2y' + y = 0

the characteristic equation has repeated roots at r = -1

So the homogeneous solution is:

y[sub:bup7lg6z]h[/sub:bup7lg6z] = K[sub:bup7lg6z]1[/sub:bup7lg6z] * e[sup:bup7lg6z]-x[/sup:bup7lg6z] + K[sub:bup7lg6z]2[/sub:bup7lg6z] * x * e[sup:bup7lg6z]-x[/sup:bup7lg6z]

The particular solution is

y[sub:bup7lg6z]p[/sub:bup7lg6z] = A * x[sup:bup7lg6z]2[/sup:bup7lg6z] + B*x + C

Using these into original ODE we get

A * x[sup:bup7lg6z]2[/sup:bup7lg6z] + (4*A + B)*x + (2*A + 2*B + C) = x[sup:bup7lg6z]2[/sup:bup7lg6z] + 2

Solve for A, B & C

Then your solution is:

y = y[sub:bup7lg6z]h[/sub:bup7lg6z] + y[sub:bup7lg6z]p[/sub:bup7lg6z] = K[sub:bup7lg6z]1[/sub:bup7lg6z] * e[sup:bup7lg6z]-x[/sup:bup7lg6z] + K[sub:bup7lg6z]2[/sub:bup7lg6z] * x * e[sup:bup7lg6z]-x[/sup:bup7lg6z] + A * x[sup:bup7lg6z]2[/sup:bup7lg6z] + B*x + C

Now solve for K[sub:bup7lg6z]1[/sub:bup7lg6z] & K[sub:bup7lg6z]2[/sub:bup7lg6z] using the initial conditions.

 

[c4l3b]

To find particular solution
y'' + 2y' + y = x^2 + 2

conditions y(0) = 3 and y(0) = 1

Since this a in-homogeneous equation do I just solve the RHS seperately x^2 and 2?
Would I need to express finding the particular function?

----------------------------------------------------------------------------------------

y + y' = 3

x > = 0

conditions y(o) = 1


y = A cos x + B sin x <<< Incorrect - Does it satisfy your given ODE (y" + 2y + y = x[sup:trm566lz]2[/sup:trm566lz] + 2)?

y' = -A sin x + B cos x and y'' = -A cos x - B sin x

I know this is the solution but its those damn conditions again, I mean if zero was replace with pi what would happen?

Advice please.

y'' + 2y' + y = 0

the characteristic equation has repeated roots at r = -1

So the homogeneous solution is:

y[sub:trm566lz]h[/sub:trm566lz] = K[sub:trm566lz]1[/sub:trm566lz] * e[sup:trm566lz]-x[/sup:trm566lz] + K[sub:trm566lz]2[/sub:trm566lz] * x * e[sup:trm566lz]-x[/sup:trm566lz]

The particular solution is

y[sub:trm566lz]p[/sub:trm566lz] = A * x[sup:trm566lz]2[/sup:trm566lz] + B*x + C

Using these into original ODE we get

A * x[sup:trm566lz]2[/sup:trm566lz] + (4*A + B)*x + (2*A + 2*B + C) = x[sup:trm566lz]2[/sup:trm566lz] + 2

Solve for A, B & C

Then your solution is:

y = y[sub:trm566lz]h[/sub:trm566lz] + y[sub:trm566lz]p[/sub:trm566lz] = K[sub:trm566lz]1[/sub:trm566lz] * e[sup:trm566lz]-x[/sup:trm566lz] + K[sub:trm566lz]2[/sub:trm566lz] * x * e[sup:trm566lz]-x[/sup:trm566lz] + A * x[sup:trm566lz]2[/sup:trm566lz] + B*x + C

Now solve for K[sub:trm566lz]1[/sub:trm566lz] & K[sub:trm566lz]2[/sub:trm566lz] using the initial conditions.

y = A cos x + B sin x <<< Incorrect - Does it satisfy your given ODE (y" + 2y + y = x[sup:trm566lz]2[/sup:trm566lz] + 2)?

No its for;

y + y' = 3

x > = 0

conditions y(o) = 1

 

[Deleted member 4993]

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y + y' = 3

x > = 0

conditions y(o) = 1


y = A cos x + B sin x <<< Incorrect - Does it satisfy your given ODE (y" + 2y + y = x[sup:3hhd54r7]2[/sup:3hhd54r7] + 2)?

No its for; <<< Ask the same question - Does it satisfy your given ODE (y + y' = 3)?


y + y' = 3

x > = 0

conditions y(o) = 1