seperation of variables to dolve differential equasions

[kris11111]

i have these 3 questions in which i have no idea how to do

use seperation of variable to solve the following differential equasions

a) dy/dx=5-x/y^2

b) y'=y^2-e^3t y^2, y(0)=1 (note 3t is to the power of)

c) y'+y=1, y(0)=2


how do i do these? any help would be awesome.

 

[tkhunny]

a) dy/dx=5-x/y^2

Fix your notation. Are you SURE you don't mean dy/dx = (5-x)/y^2 ?

Separate: y^2 dy = (5-x) dx

That's all there is to it.

 

[kris11111]

no whats written is correct there is no brackets

 

[galactus]

i have these 3 questions in which i have no idea how to do

use seperation of variable to solve the following differential equations

a) \(\displaystyle \frac{dy}{dx}=5-\frac{x}{y^{2}}\)

If you are sure this is the correct problem, then there must be a typo in the text. This is not doable by separating variables.

It requires more advanced methods. If it were written as tkh suggested, then no problem.

b) \(\displaystyle \frac{dy}{dt}=y^{2}-e^{3t} y^{2}, \;\ y(0)=1\)

Factor the right side and separate variables:

\(\displaystyle \frac{dy}{dt}=y^{2}(1-e^{3t})\)

Separate:

\(\displaystyle \frac{dy}{y^{2}}=(1-e^{3t})dt\)

Now, integrate and continue.

c). \(\displaystyle \frac{dy}{dt}+y=1, \;\ y(0)=2\)

Separate \(\displaystyle \frac{dy}{1-y}=dt\)

Now, integrate and continue.

 

[tkhunny]

no whats written is correct there is no brackets

Very irritating. Sorry that so many who do it badly make it hard to translate those who do it well.

 

[kris11111]

so for
b. i get x(dy/Dt=Dt(1-e^3t)
c. i get x(Dt+y=dy)

are these right

and a. ??????? lol

 

[Deleted member 4993]

so for
b. i get x(dy/Dt=Dt(1-e^3t)
c. i get x(Dt+y=dy)

are these right

and a. ??????? lol

No....

You are being very careless ... thus your answers do not make sense ( you are mixing up independent variable 'x' with 't' - and vice versa)

It does not look like you have studied these at all ... please review some worked out problems in your textbook

 

[kris11111]

its my first attempt and im useless at math with letters. hense mixing up variables

 

[tkhunny]

its my first attempt and im useless at math with letters. hense mixing up variables

Wow! Why on Earth are you in Differential Equations? This makes no sense at all. Please reconsider something - goals, academic pursuits, attitude - something isn't right, here.

 

[kris11111]

unfortunatly my uni requires me to do this math even though i will never use it in my career its just one of those things that are an obsticle to what i want to be yet has no relevence in my future. quiting is not an option, just need help learning

 

[tkhunny]

Good luck with that. Seems like you missed a few things along the way. In particular, you seem to have missed that much of the learning of mathematics is learning how to think in a rational and organized manner. You WILL use that in your future - whatever your primary field. Maybe you could focus on that concept, rather than the useless idea that somehow memorizing a few facts and techniques is supposed to do you any good.

 

[galactus]

so for
b. i get x(dy/Dt=Dt(1-e^3t)
c. i get x(Dt+y=dy)

are these right

and a. ??????? lol

I am sorry, but I do not know what this is.

I will do part b, then you show me the other.

After separating, we have \(\displaystyle \frac{dy}{y^{2}}=(1-e^{3t})dt\)

Integrate both sides:

\(\displaystyle \int \frac{1}{y^{2}}dy=\int (1-e^{3t})dt\)

\(\displaystyle \frac{-1}{y}=t-\frac{1}{3}e^{3t}+C\)

Solve for y:

\(\displaystyle y=\frac{3}{e^{3t}-3(t+C)}\)

Now, if there were initial conditions, we could use them to find C.

So, knowing that y=1 when t=0, we sub them in and solve for C:

\(\displaystyle 1=\frac{3}{e^{3(0)}-3(0+C)}\)

\(\displaystyle 1=\frac{3}{1-3C}\)

\(\displaystyle C=\frac{-2}{3}\)

So, the final solution ends up being:

\(\displaystyle y=\frac{3}{e^{3t}-3(t-\frac{2}{3})}\)

Now, follow the same method for the other and let me know what you get.

tkh is a good guy. He is not being rude, he is just trying to get you to write mathematically clear so that other know what you mean.

How would your instructor respond to the solutions you posted if you were to hand these in on a test or homework?. See how I done this one. Be detailed and clear.