wondering if someone could give me a hand on these couple of questions

1. (x^2+1)dy/dx+y^2+1=0 y(0)=1

2. (x^2-x-2)dy/dx-y=0

3. dy/dx-y=3e^x

Any help would be appreciated

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- Aug 21st 2006, 02:12 PMaction259homework help - differential equations
wondering if someone could give me a hand on these couple of questions

1. (x^2+1)dy/dx+y^2+1=0 y(0)=1

2. (x^2-x-2)dy/dx-y=0

3. dy/dx-y=3e^x

Any help would be appreciated - Aug 21st 2006, 02:18 PMJamesonQuote:

Originally Posted by**action259**

Rewrite:

Those are both arctan integrals so you can take it from there! - Aug 21st 2006, 02:24 PMJamesonQuote:

Originally Posted by**action259**

Rewrite:

The left integral is trivial but for the second one use partial fractions.

Now:

Let x=1, so B=-1

Let x=2

So - Aug 21st 2006, 03:39 PMtopsquark
[QUOTE=action259]dy/dx-y=3e^x[\QUOTE]

First solve the homogeneous equation:

Solve the (characteristic?) equation .

So m = 1. Thus .

Now we need a particular solution. Since the RHS of the original equation is , try:

( is a solution to the homogeneous equation so it won't work here.)

so inserting these into the original equation we get:

Thus B = 3.

So the final solution is .

-Dan - Aug 21st 2006, 03:53 PMThePerfectHackerQuote:

Originally Posted by**action259**

---

Regocnize this to be a*linear-first order differential equation*.

Here,

and

Thus,

And,

Thus, the solution(s) is,

Thus,

- Aug 21st 2006, 04:22 PMThePerfectHacker
Method #3

---

Express as,

Thus,

(1)

Find the integrating factor,

Multiply both sides by it,

Now check if it is conservative,

---Good.

We have turned this differencial equation into an*exact differenciatial equation*.

Therefore, there exists a function,

such as,

gives our equation above.

---

Let be such a function. Then,

Thus,

Solving the second partial differential equation yields,

(a function of x alone).

Thus,

Thus,

Thus,

Therefore, the function is,

Thus,

Thus,

- Aug 21st 2006, 05:11 PMaction259
thanks for the help it was really useful...i got another question. I'm doing a question that asks me to find the value of the family of curves if the slope of the tangent at any point is (1-2x)/(1+y)

now does this mean that dy/dx=(1-2x)/(1+y), which is a separable equation, i'm not really sure what i am supose to find. if i solve the equation i get

x-x^2=y+y^2, how does this help me? - Aug 21st 2006, 07:26 PMThePerfectHackerQuote:

Originally Posted by**action259**

Which is seperable. - Aug 21st 2006, 07:45 PMSoroban
Hello, action259!

Check the original wording of the problem, please.

There is no "*value*of the family of curves".

Quote:

Find the family of curves if the slope of the tangent at any point is:

Does this mean that , which is a separable equation? . . . yes!

If i solve the equation i get: . . . um, not quite

How does this help me?

Separate the variables: .

Integrate: .

We have:

. . . .

Complete the square:

. . . .

And we have:

. . . . . . .

The family is the set of all circles with center at

. . and radius . . . where