I need to solve the following differential equation

Kale Bright

Kale Bright

Answered question

2022-04-06

I need to solve the following differential equation with initial condition y(0)=1
x2(2y+y2)dx=2x3dy

Answer & Explanation

glanzerjbdo

glanzerjbdo

Beginner2022-04-07Added 13 answers

Step 1
You should rewrite the equation as y(x)2+2y(x)=2xy′(x). Notice that y2+2y=y(y+2), so what you have is y(x)=−2 and y(x)=0 as special solutions, though these do not satisfy your initial condition. To obtain the other solutions, you have that
1=2xy′(x)y(x)[y(x)+2]
=([y(x)+2]−y(x))xy′(x)y(x)[y(x)+2]
=xy′(x)y(x)−xy′(x)y(x)+2.
The problem here is that this clearly demonstrates this differential equation is incompatible with the condition y(0)=1. This is because it implies
1=2⋅0⋅{y′(0)}y(0)[y(0)+2]=0⋅{y′(0)}3=0y′(0),
which is impossible. What this suggests is that y is not differentiable at 0, but satisfies the differential equation everywhere in R{0}. In that case, though, the initial condition given is irrelevant, as it does not constrain the set of solutions at all.
The equation can be written as
y′(x)y(x)−y′(x)y(x)+2=1x.
There are three families of solutions to consider here. One family of solutions is when y(x)<−2, in which case, the differential equation implies
ln⁡[−y(x)]−ln⁡[−y(x)−2]=ln⁡(−x)+A,,x<0
ln⁡[−y(x)]−ln⁡[−y(x)−2]=ln⁡(x)+B,,x>0
which is equivalent to
y(x)y(x)+2=−eAx,,x<0
y(x)y(x)+2=eBx,,x>0
which is equivalent to
y(x)=−2xe−A+x,,x<0
y(x)=2xe−B−x,,x>0.
Another family of solutions to consider is when −2<{y(x)}<0,, in which case the differential equation implies
ln⁡[−y(x)]−ln⁡[y(x)+2]=ln⁡(−x)+A,,x<0
ln⁡[−y(x)]−ln⁡[y(x)+2]=ln⁡(x)+B,,x>0
which is equivalent to
y(x)y(x)+2=eAx,,x<0
y(x)y(x)+2=−eBx,,x>0
which is equivalent to
y(x)=2xe−A−x,,x<0
y(x)=−2xe−B+x,,x>0
The first family of solutions is the same family of solutions as when y(x)>0, and also, that family only works on the region (−e−A,e−B)∖{0}.. Outside that region, one can have the second family, or one of the constant solutions. However, there is no family of solutions where y(0)=1 is allowed.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?

Product

Company

Connect with us

Get Plainmath App

Plainmath Logo

E-mail us: [email protected]

Our Service is useful for:

Plainmath is a platform aimed to help users to understand how to solve math problems by providing accumulated knowledge on different topics and accessible examples.

2023 Plainmath. All rights reserved