Find the solution of the following Differential Equations y'= (y-x)

reproacht3

reproacht3

Answered question

2021-12-26

Find the solution of the following Differential Equations
y=(yx)

Answer & Explanation

puhnut1m

puhnut1m

Beginner2021-12-27Added 33 answers

dydx=(yx)2
Let yx=t
dydx1=dtdx
dydx=dtdx+1
dtdx+1=t2
dtdx=t21
dtt21=dx
12ln(t1t+1)=x+c
ln(yx1yx+1)=2x+2c
Jimmy Macias

Jimmy Macias

Beginner2021-12-28Added 30 answers

y=x+a2+x2a2x2
where a is a constant of integration.
Explanation:
Let us change variables to u=yx
Then u=y1 and so the differential equation becomes
u=u21duu21=dx12log|u1u+1|=logxloga
where we have written the constant of integration as loga. Thus u1u+1=x2a2u=a2+x2a2x2
karton

karton

Expert2022-01-09Added 613 answers

Explanation:
We have: y=(yx)2[A]
Perform the substitution:
u=y-x
Then differentiating wrt x, and applying the product rule:
dudx=dydx1dydx=1+dudx
Substituting into the DE:
1+dudx=u2
dudx=u21
Reducing the DE to a separable form:
1u21dudx=1
So we can ''separate the variables'' to get:
1u21du=dx
We can perform partial fraction decomposition of the LHS integrand:
1u21=1(u+1)(u1)
=Au+1+Bu1
And using the ''cover up'' methods we get:
A=12,B=12
Allowing us to write the equation as:
121u11u+1du=dx
And integrating, we get:
12ln(u1)ln(u+1)=x+Cln(u1u+1)=2x+2Cu1u+1=e2x+2Cu1=(u+1)Ae2xu1=uAe2x+Ae2xuuAe2x=1+Ae2xu(1Ae2x)=1+Ae2xu=1+Ae2x1Ae2x

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