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, 01:12 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, 01:18 PMJamesonQuote:

Originally Posted by**action259**

$\displaystyle (x^2+1)dy=(-y^2-1)dx$

Rewrite: $\displaystyle -\frac{1}{y^2+1}dy=\frac{1}{x^2+1}dx$

Those are both arctan integrals so you can take it from there! - Aug 21st 2006, 01:24 PMJamesonQuote:

Originally Posted by**action259**

$\displaystyle (x^2-x-2)dy=ydx$

Rewrite: $\displaystyle \frac{1}{y}dy=\frac{1}{x^2-x-2}dx$

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

$\displaystyle \frac{1}{x^2-x-2}=\frac{1}{(x-2)(x+1)}$

Now: $\displaystyle \frac{A}{x-2}+\frac{B}{x+2}=\frac{1}{x^2-x-2}$

$\displaystyle A(x-1)+B(x-2)=1$

Let x=1, $\displaystyle -B=1$ so B=-1

Let x=2 $\displaystyle A=1$

So $\displaystyle \frac{1}{y}dy=\frac{1}{x-2}-\frac{1}{x+1}dx$ - Aug 21st 2006, 02:39 PMtopsquark
[QUOTE=action259]dy/dx-y=3e^x[\QUOTE]

First solve the homogeneous equation:

$\displaystyle \frac{dy_h}{dx} - y_h = 0$

Solve the (characteristic?) equation $\displaystyle m - 1 = 0$.

So m = 1. Thus $\displaystyle y_h = Ae^{x}$.

Now we need a particular solution. Since the RHS of the original equation is $\displaystyle e^x$, try:

$\displaystyle y_p = Bxe^x$ ($\displaystyle e^x$ is a solution to the homogeneous equation so it won't work here.)

$\displaystyle \frac{dy_p}{dx} = Be^x + Bxe^x$

so inserting these into the original equation we get:

$\displaystyle Be^x + Bxe^x - Bxe^x = 3e^x$

Thus B = 3.

So the final solution is $\displaystyle y = y_h + y_p = Ae^x + 3xe^x$.

-Dan - Aug 21st 2006, 02:53 PMThePerfectHackerQuote:

Originally Posted by**action259**

---

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

Here,

$\displaystyle P(x)=-1$ and $\displaystyle Q(x)=3e^x$

Thus,

$\displaystyle I(x)=e^{\int -1 dx}=e^{-x}=e^{-x}$

And,

$\displaystyle \int I(x)Q(x)dx=\int 3e^{-x}e^{x}dx=3x+C$

Thus, the solution(s) is,

$\displaystyle \frac{1}{e^{-x}}(3x+C)$

Thus,

$\displaystyle y=3xe^x+Ce^x$ - Aug 21st 2006, 03:22 PMThePerfectHacker
Method #3

---

Express as,

$\displaystyle dy-ydx=3e^xdx$

Thus,

$\displaystyle (-y-3e^x)dx+(1)dy=0$ (1)

Find the integrating factor,

$\displaystyle e^{-x}$

Multiply both sides by it,

$\displaystyle (-ye^{-x}-3)dx+(e^{-x})dy=0$

Now check if it is conservative,

$\displaystyle \frac{\partial (-ye^{-x}-3)}{\partial y}=\frac{\partial (e^{-x})}{\partial x}\rightarrow -e^{-x}=-e^{-x}$---Good.

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

Therefore, there exists a function,

$\displaystyle f(x,y)=K$ such as,

$\displaystyle d(f(x,y))=0$ gives our equation above.

---

Let $\displaystyle f(x,y)$ be such a function. Then,

$\displaystyle d(f(x,y))=f_x(x,y)dx+f_y(x,y)dy$

Thus,

$\displaystyle f_x(x,y)=-ye^{-x}-3$

$\displaystyle f_y(x,y)=e^{-x}$

Solving the second partial differential equation yields,

$\displaystyle f(x,y)=ye^{-x}+\phi(x)$ (a function of x alone).

Thus,

$\displaystyle f_x(x,y)=-ye^{-x}+\phi'(x)$

Thus,

$\displaystyle \phi'(x)=-3$

Thus,

$\displaystyle \phi(x)=-3x+C$

Therefore, the function is,

$\displaystyle f(x,y)=-ye^{-x}-3x+C$

Thus,

$\displaystyle -ye^{-x}-3x+C=K$

Thus,

$\displaystyle y=3xe^x+Ce^x$ - Aug 21st 2006, 04:11 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, 06:26 PMThePerfectHackerQuote:

Originally Posted by**action259**

$\displaystyle \frac{dy}{dx}=\frac{1-2x}{1+y}$

Which is seperable. - Aug 21st 2006, 06:45 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: $\displaystyle \frac{1-2x}{1+y}$

Does this mean that $\displaystyle \frac{dy}{dx} \:=\:\frac{1-2x}{1+y}$, which is a separable equation? . . . yes!

If i solve the equation i get: $\displaystyle x - x^2\:=\:y+y^2$ . . . um, not quite

How does this help me?

Separate the variables: .$\displaystyle (1 + y)\,dy \:=\:(1 - 2x)\,dx$

Integrate: .$\displaystyle y + \frac{1}{2}y^2\;=\;x - x^2 + C$

We have:

. . . . $\displaystyle x^2 - x \qquad + y^2 + \frac{1}{2}y \qquad \;=\;C$

Complete the square:

. . . . $\displaystyle x^2 - x + \frac{1}{4} + y^2 + \frac{1}{2}y + \frac{1}{16}\;=\:C + \frac{1}{4} + \frac{1}{16} $

And we have:

. . . . . . . $\displaystyle \left(x - \frac{1}{2}\right)^2 + \left(y + \frac{1}{4}\right)^2\;=\;C + \frac{5}{16} $

The family is the set of all circles with center at $\displaystyle \left(\frac{1}{2},-\frac{1}{4}\right) $

. . and radius $\displaystyle \sqrt{C + \frac{5}{16}}$ . . . where $\displaystyle C \geq -\frac{5}{16}$