Starting Differential Equations: Finding the General Solution

[UCN Student]

I am having trouble starting differential equations it says to find the general solution of such and i don't know where to get started on some of the.

examples:

dR
--- = tan0 0=theta
d0

dy
--- = 3x- 3y
dx

i don't want them answered as they are part of my assignment i just want help on how to go about starting to solve them.

thank you

Ryan

 

[PrinceOfDarkness]

A solution to a DE means that the value of 'x' or whatever the variable is, satisfies the equation. There can be infinitely many solutions to a DE!

You should better consult your textbook. Or read Schaum's outline of DEs. I don't think anyone will solve these Qs here for you. We need to know that at least you tried.

Hint: Separate Variables and Integrate!

 

[HallsofIvy]

For the first one, rewrite it as
$$dR= tan(\theta)d\theta$$
and integrate both sides.

The second one is a "linear, first order" differential equation and I'll bet your textbook has some detailed information about those!

 

[UCN Student]

ok thanks a lot that helps me out a lot.

 

[UCN Student]

so for the dR = tan(0)d0

would the answer be:

y=-ln cos0+c

 

[UCN Student]

and for:

dy
-- = 3x-3y
dx

would the answer be:

y= ((3x^4)/4) - ((3y^4)/4)

 

[FrogPad]

You can check your answer.

for example)

$$\frac{dy}{dx} = 3x-3y$$
This is saying that, a function exists $y$ that when you differentiate it with respect to $x$ then it is equal to $3x-3y$

So how can you check your answer?

Well your answer is saying that

$$y= \frac{1}{4}3x^4 - \frac{1}{4}3y^4$$

So if you differentiate your function.

$$\frac{dy}{dx} = ?$$

Is that differentiated function equal to the right hand side (the $3x-3y$)?

Also what happened to the $c$ (don't forget the constant of integration) when you integrated? A general solution will have infinitely many solutions, so that $c$ is important. Otherwise it is not a general solution.