another prob on 3rd order differential eqns.

[Lost souls]

(D-1)³y=16e^3x​

taking the auxiliary function,
(D-1)³=0
D=1,1,1

the particular integral would be (1/f(a))e^ax


(D³-3D²+3D-1)^-1e^3x = 16e^3x(1/8) = 2e^3x

complete solution=> y = C.F + 2e^3x

what will be the complementary function??

 

[Deleted member 4993]

(D-1)³y=16e^3x​

taking the auxiliary function,
(D-1)³=0
D=1,1,1

the particular integral would be (1/f(a))e^ax


(D³-3D²+3D-1)^-1e^3x = 16e^3x(1/8) = 2e^3x

complete solution=> y = C.F + 2e^3x

what will be the complementary function??

Since your homogeneous solution and the forcing function (RHS) are equal - you have resonance effect.

How do you solve those types for 2nd. order ODE?

Follow similar route....

 

[Lost souls]

Since your homogeneous solution and the forcing function (RHS) are equal - you have resonance effect.

How do you solve those types for 2nd. order ODE?

Follow similar route....

sorry. .you lost me at resonance effect. .btw, for 2nd order DE with real and same roots, y=(C₁+C₂x)e^ax

 

[Deleted member 4993]

sorry. .you lost me at resonance effect. .btw, for 2nd order DE with real and same roots, y=(C₁+C₂x)e^ax

Thus your homogeneous solution should be - as the roots are repeated 3 times:

y_H = (A + Bx + Cx²) * e^3x

and

y_P= Kx³ * e^3x

Where K will be evaluated by equating the coefficients after applying the original ODE.

Just like the second order ODE (it will be messy algebra).

 

[Lost souls]

Thus your homogeneous solution should be - as the roots are repeated 3 times:

y_H = (A + Bx + Cx²) * e^3x

and

y_P= Dx³ * e^3x

Where D will be evaluated by equating the coefficients after applying the original ODE.

Just like the second order ODE (it will be messy algebra).

what is y_p?? isnt the soln y= (a + bx + cx²)*e^{3x +} 2e^{​3x ??}

 

[Deleted member 4993]

what is y_p?? isnt the soln y= (a + bx + cx²)*e^3x + 2e^{​3x ??}

Did you apply your proposed solution to your ODE and see if it is satisfied?

 

[HallsofIvy]

Since your homogeneous solution and the forcing function (RHS) are equal - you have resonance effect.

How do you solve those types for 2nd. order ODE?

Follow similar route....

??? There is no "resonance". The characteristic roots are 1, 1, 1, so the independent solutions to the homogeneous equation are \(\displaystyle e^x\), \(\displaystyle xe^x\), and \(\displaystyle x^2e^x\). The forcing function is \(\displaystyle e^{3x}\), completely different.

 

[Deleted member 4993]

??? There is no "resonance". The characteristic roots are 1, 1, 1, so the independent solutions to the homogeneous equation are \(\displaystyle e^x\), \(\displaystyle xe^x\), and \(\displaystyle x^2e^x\). The forcing function is \(\displaystyle e^3x\), completely different.

You are very correct .... I wasn't paying attention to the whole problem - I took it that e^3xis a part of the homogeneous solution (instead of e^x)

Sorry about that....

 

[Lost souls]

You are very correct .... I wasn't paying attention to the whole problem - I took it that e^3xis a part of the homogeneous solution (instead of e^x)

Sorry about that....

so. . y= (a + bx + cx²)*e^x + 2e^​3x is correct, innit?

 

[HallsofIvy]

so. . y= (a + bx + cx²)*e^x + 2e^​3x is correct, innit?

Yes, that is correct.

 

[Deleted member 4993]

However, you should make sure that - the proposed solution does satisfy the given ODE.

 

[Lost souls]

i'll do that. . .thanx again, both of you, HallsofIvy and subhotosh. .