What is a solution to the differential equation dy/dy=√1−y?

rustenig 2022-09-14 Answered
What is a solution to the differential equation d y d y = 1 - y ?
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Answers (1)

Koen Henson
Answered 2022-09-15 Author has 17 answers
If this is written correctly, then you know that d y d y = 1 . That makes this really easy. But first, let's write this in the standard form:
d y d y - 1 - y = 0
Now, let's use the fact that d y d y = 1 to get:
1 - 1 - y = 0
1 - ( 1 - y ) = 0
y = 0

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asked 2022-06-29
I have been trying to solve a differential equation as a practice question for my test, but I am just unable to get the correct answer. Please have a look at the D.E:
d y / d x = 1 / ( 3 x + sin ( 3 y ) )
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d x / d y = 3 x + sin ( 3 y )
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x ( 3 / 2 ) x 2 = ( 1 / 3 ) cos ( 3 y ) + c
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asked 2022-06-22
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( s e c x   y ) = s i n 2 x s e c 2 x
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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:
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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
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asked 2022-07-02
I'm trying to understand what is wrong with this solution, since I'm not getting the same answer in Matlab
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This is what I'm getting in Matlab
( ( tanh ( x 2 8 + atanh ( 3 ) 1 8 ) + 1 ) 2 4 ( tanh ( x 2 8 + atanh ( 5 ) + 1 8 ) 1 ) 2 4 )

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