D operator - Differential equations

Scary

hello can anyone help me with this problem just parts (ii) thanks, the rest i've managed to solve

thanks again

Scroll to see replies

Reply 1

Math12345

What have you tried ? What is equation 1? How did you solve the others parts?

Reply 2

Scary

the first part of the question was to use the auxiliary equation to find the yc and then finding the particular integral, the equation was d^2y/dx^2 -dy/dx -6y = e^-2x, the general solution i got was -1/5xe^(-2x) +Ae^(3x)+Be^(-2x)

Reply 3

simon0

The general solution is correct.

Make the substitution suggested:

Y = (D+2)y

Therefore:

(D-3)Y = e^{-2x}

\Rightarrow \frac{dY}{dx} - 3Y = e^{-2x}

.

Can you see where to go from here?

(edited 7 years ago)

Reply 4

Scary

Original post by simon0

The general solution is correct.

Make the substitution suggested:

Y = (D+2)y

Therefore:

(D-3)Y = e^{-2x}

\Rightarrow \frac{dY}{dx} - 3Y = e^{-2x}

.

Can you see where to go from here?

hi, thanks for the help, would you then solve the equation you put as separable or would you need to use an integrating factor?

Reply 5

Zacken

Original post by Scary

hi, thanks for the help, would you then solve the equation you put as separable or would you need to use an integrating factor?

Integrating factor, at a glance.

Reply 6

simon0

Original post by Scary

hi, thanks for the help, would you then solve the equation you put as separable or would you need to use an integrating factor?

You can try placing the formula in a separable form!

You will need to use the integrating factor method to solve the first order ordinary diferential equation.

Reply 7

Scary

Original post by simon0

You can try placing the formula in a separable form!

You will need to use the integrating factor method to solve the first order ordinary diferential equation.

okay thanks, the solution i got to was y=ce^3x-(e^-2x)/5 is this correct?

Reply 8

username2769500

Original post by Scary

hello can anyone help me with this problem just parts (ii) thanks, the rest i've managed to solve

thanks again

Differentiate rhs >> e-2x/-2x
Which is D
Sub into D-3
Let Y = the other bit or equal rhs/(D-3)

... Keep playing about with that until you find the answer.

Reply 9

simon0

Original post by Scary

okay thanks, the solution i got to was y=ce^3x-(e^-2x)/5 is this correct?

Nearly...

Replace

y

with

Y

.

Now you should have a formula equal to

Y

which is a substitution for

Dy + 2y

.

Do you know what to do after?

Reply 10

simon0

Original post by Anfanny

Differentiate rhs >> e-2x/-2x
Which is D...

How did you differentiate the RHS?

Also,

D

is an operator that operates on the term to the immediate right of it.

(edited 7 years ago)

Reply 11

username2769500

D = d/dx which means the function. We are told that in terms of D LHS is same as the exponent e-2x. The de/dx is the differential and the rule is divide by the power constant (-2) for exponents.
You must learn how to figure out the maths language so you can read the maths. Do a lot of c3/c4 again from various exam boards to "learn this language" i.e spotting that the left hand side is the function in terms of D and y and the rhs is the function in terms of exponents. The substitution gives an identity which we can follow the identity rule to solve. Hope this helps.

Reply 12

username2769500

Original post by simon0

How did you differentiate the RHS?

Also,

D

is an operator that operates on the term to the immediate right of it.

D is a value and we are told its d/dx it can't be the LHS as we are told D equals it therefore it's the right hand side. Or that/y as then no d will exist on the tha in these forms.

Reply 13

simon0

Then why did you say the derivative of the RHS (with respesct to

x

) is e-2x/-2x?

Reply 14

username2769500

Is that not the rule for exponents? Sorry well it's either multiply, decide or just ln(x). The right hand side differentiated is D. I think. as it has X in it... The LHS does not therefore it is f(X) and the LHS is the f(y).