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# D operator - Differential equations

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1. hello can anyone help me with this problem just parts (ii) thanks, the rest i've managed to solve thanks again
2. What have you tried ? What is equation 1? How did you solve the others parts?
3. 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)
4. The general solution is correct.

Make the substitution suggested:

Therefore:

.

Can you see where to go from here?
5. (Original post by simon0)
The general solution is correct.

Make the substitution suggested:

Therefore:

.

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?
6. (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.
7. (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.
8. (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?
9. (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)

10. (Original post by Scary)
okay thanks, the solution i got to was y=ce^3x-(e^-2x)/5 is this correct?
Nearly...

Replace with .

Now you should have a formula equal to which is a substitution for .

Do you know what to do after?
11. (Original post by Anfanny)
Differentiate rhs >> e-2x/-2x
Which is D...
How did you differentiate the RHS?

Also, is an operator that operates on the term to the immediate right of it.
12. 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.
13. 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.

(Original post by simon0)
How did you differentiate the RHS?

Also, 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.
14. Then why did you say the derivative of the RHS (with respesct to ) is e-2x/-2x?
15. 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).
16. (Original post by Anfanny)
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).
Hi, for the case of , the derivative is obtained using the chain rule.

,

,

,

.
17. Also:
is an operator (or functional) where it operates on whatever is multiplied to it on the right.

Therefore:

The RHS is NOT and therefore is not the RHS differentiated.

If I have misunderstood you, I do apologise.
18. (Original post by simon0)
Nearly...

Replace with .

Now you should have a formula equal to which is a substitution for .

Do you know what to do after?
So would it be dy/dx +2y=rhs and you can solve this using an intergration factor ?
19. Yes, spot on.
20. okay thank you, could you provide me with the general solution so i can work towards it, its causing me some problems
(Original post by simon0)
Yes, spot on.

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