How do you find all solutions of the differential equation d^2y/dx^2=0?

calcific5z 2022-09-11 Answered
How do you find all solutions of the differential equation d 2 y d x 2 = 0 ?
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Answers (1)

Peyton Cox
Answered 2022-09-12 Author has 18 answers
This is a Second Order homogeneous DE with constant coefficients, but separable so we can just integrate (twice):
d 2 y d x 2 = 0
Integrate:
d y d x = A
Integrate again:
y = A x + B
which is the General Solution
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I would like to solve:
y y x y 2 = 0
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y = x 2 4 + C
Now, if we want to verify the solution, it turns out that C must be zero, in other words, y = x 2 / 4 satisfies the original differential equation.
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2) If we try to solve this differential equation with Mathematica, we obtain
y = C 1 x + C 1 2 ,
which has a different form from the analytical approach. How can we also produce this result analytically?
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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?