Substitution in Differential Equations (still need help)

[Weatherkid11]

Find the general solution of the differential equation (x+y)y'=1
First I made the DE equal to y'--> y'=1/(x+y).
Then I got my v--> v=x+y, then y=v-x so y'=v'-1
From here, I do not know how to finish. Please help. Thanks!

 

[Daniel_Feldman]

Re: Substitution in Differential Equations

Find the general solution of the differential equation (x+y)y'=1
First I made the DE equal to y'--> y'=1/(x+y).
Then I got my v--> v=x+y, then y=v-x so y'=v'-1
From here, I do not know how to finish. Please help. Thanks!

You're not quite taking the right approach here...

Differential equations: First you separate variables>>get x's and dx's on one side and y's and dy's on the other (recall: y'=dy/dx).

The integrate both sides, and you should come up with a solution. Don't forget the constant of integration!

 

[Weatherkid11]

Yes but arent you supposed to use substitution since the topic of the chapter is Substitution Methods and Exact Equations, then after you substitute you separate the variables?

 

[Daniel_Feldman]

Yes but arent you supposed to use substitution since the topic of the chapter is Substitution Methods and Exact Equations, then after you substitute you separate the variables?

Hmmm...I'm not sure. I've always solved differential equations by going straight to separation of variables.


Let's let someone else have a look at this and see what they think.

 

[Unco]

I would solve the DE by using the integrating factor \(\displaystyle \mbox{\mu = e^{-y}}\) to make it exact.

 

[soroban]

Hello, Weatherkid11!

Find the general solution of the differential equation $\displaystyle (x\,+\,y)\frac{dy}{dx}\:=\:1$

First I made the DE equal to $\displaystyle \frac{dy}{dx}:\;\frac{dy}{x}\:=\:\frac{1}{x\,+\,y}$

Then I got my $\displaystyle v:\;v\:=\:x\,+\,y\;\;\Rightarrow\;\;y\:=\:v\,-\,x\;\;\Rightarrow\;\;\frac{dy}{dx}\:=\:\frac{dv}{dx}\,-\,1$
From here, I do not know how to finish.

Substitute: \(\displaystyle \L\:\frac{dv}{dx}\,-\,1\:=\:\frac{1}{v}\)

Some algebra: \(\displaystyle \L\:\frac{dv}{dx}\:=\:\frac{1}{v}\,+\,1\:=\:\frac{1\,+\,v}{v}\)

Separate variables: $\displaystyle \:\frac{v}{1\,+\,v}\,dv\:=\:dx\;\;\Rightarrow\;\;\left(1\,-\,\frac{1}{v\,+\,1}\right)\,dv\:=\:dx$

Integrate: $\displaystyle \:v - \ln|v\,+\,1|\:=\:x\,+\,c$

Back-substitute: \(\displaystyle \x\,+\,y) - \ln|x\,+\,y\,+\,1|\:=\:x\,+\,c\)

Simplify: $\displaystyle \:\ln|x\,+\,y\,+\,1|\:=\:y\,+\,c$

Then: $\displaystyle \:x\,+\,y\,+\,1\:=\:e^{y+c} \:=\:e^y\,\cdot\,e^c$

Therefore: $\displaystyle \;x\,+\,y\,+\,1\:=\:Ce^y$