How do you find all solutions of the differential equation d^2y/dx^2=x^(−2)?

frobirrimupyx 2022-09-13 Answered
How do you find all solutions of the differential equation d 2 y d x 2 = x - 2 ?
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Answers (1)

Marie Horn
Answered 2022-09-14 Author has 12 answers
Note that when we integrate things like d x 2 we will be left with dx, so we will need to integrate this twice.
Separating variables:
d 2 y = x - 2 d x 2
Integrating:
d 2 y = x - 2 d x 2
This is analogous to doing d y = x - 2 d x y = - x - 1 + C 1 , but a dx term will still be left.
d y = ( - x - 1 + C 1 ) d x
Integrating again:
d y = ( - 1 x + C 1 ) d x
Add another constant:
y = - ln ( | x | ) + C 1 x + C 2

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asked 2022-04-10
Problem:
Solve the following differential equations.
x d y d x + y = y 2
Answer:
To solve this equation, we reduce it to a linear differential equation with the substitution v = y 3 .
x y 2 d y d x + y 3 = 1 d v d x = 3 y 2 d y d x 3 x y 2 d y d x + 3 y 3 = 3 d v d x + 3 v = 3
Now we have a first order linear differential equation. To solve it, we use the integrating factor I = e P ( x ) . In this case, we have P(x)=3.
I ( x ) = e 3 x e 3 x d v d x + 3 e 3 x v = 3 e 3 x D ( e 3 x v ) = 3 e 3 x e 3 x v = e 3 x + C v = 1 + e 3 x y 3 = 1 + e 3 x
Now to check my answer.
3 y 2 d y d x = 3 C e 3 x y 2 d y d x = C e 3 x d y d x = C e 3 x y 2 x d y d x + y = x ( C e 3 x y 2 ) + ( 1 + e 3 x ) 1 3
I cannot seem to get my answer to check. Where did I go wrong?
Here is my second attempt to solve the problem:
To solve this equation, we reduce it to a linear differential equation with the substitution v = y 3 .
x y 2 d y d x + y 3 = 1 d v d x = 3 y 2 d y d x 3 x y 2 d y d x + 3 y 3 = 3 x d v d x + 3 v = 3 d v d x + 3 x 1 v = 3 x 1
Now we have a first order linear differential equation. To solve it, we use the integrating factor I = e P ( x ) . In this case, we have P ( x ) = 3 x 1 .
I = e 3 x 1 d x = e 3 ln | x | I = 3 x 3 x d v d x + 9 v = 9
Now, I want to write:
D ( 3 x v ) = 9
but that is wrong. What did I do wrong?
Here is my third attempt to solve the problem. Last time, I made a mistake in finding the integrating factor.
To solve this equation, we reduce it to a linear differential equation with the substitution v = y 3
x y 2 d y d x + y 3 = 1 d v d x = 3 y 2 d y d x 3 x y 2 d y d x + 3 y 3 = 3 x d v d x + 3 v = 3 d v d x + 3 x 1 v = 3 x 1
Now we have a first order linear differential equation. To solve it, we use the integrating factor I = e P ( x ) . In this case, we have P ( x ) = 3 x 1 .
I = e 3 x 1 d x = e 3 ln | x | I = x 3 x 3 d v d x + 3 x 2 v = 3 x 2 D ( x 3 v ) = x 3 + C x 3 v = x 3 + C v = C x 3 + 1 y 3 = C x 3 + 1
Do I have it right now?
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I was wondering if I could get some advice on how to tackle this question:
Consider the differential equation
x 2 d y d x + 2 x y y 3 = 0 ( 3 )
Make the substitution u = y 2 and show that the differential equation reduces to
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Solve equation (4) for u(x) and hence write down the solution for equation (3).
I'm trying to do the first part of showing that the differential equation reduces to equation 4. I have started out by:
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I'm not sure where to continue on from here though.

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