# Thread: homework help - differential equations

1. ## homework 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

2. Originally Posted by action259

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

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

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

Those are both arctan integrals so you can take it from there!

3. Originally Posted by action259

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

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

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

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

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

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

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

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

Let x=2 $A=1$

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

4. [QUOTE=action259]dy/dx-y=3e^x[\QUOTE]

First solve the homogeneous equation:
$\frac{dy_h}{dx} - y_h = 0$

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

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

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

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

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

so inserting these into the original equation we get:
$Be^x + Bxe^x - Bxe^x = 3e^x$

Thus B = 3.

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

-Dan

5. Originally Posted by action259
3. dy/dx-y=3e^x
Method #2
---
Regocnize this to be a linear-first order differential equation.
Here,
$P(x)=-1$ and $Q(x)=3e^x$
Thus,
$I(x)=e^{\int -1 dx}=e^{-x}=e^{-x}$
And,
$\int I(x)Q(x)dx=\int 3e^{-x}e^{x}dx=3x+C$
Thus, the solution(s) is,
$\frac{1}{e^{-x}}(3x+C)$
Thus,
$y=3xe^x+Ce^x$

6. Method #3
---
Express as,
$dy-ydx=3e^xdx$
Thus,
$(-y-3e^x)dx+(1)dy=0$ (1)
Find the integrating factor,
$e^{-x}$
Multiply both sides by it,
$(-ye^{-x}-3)dx+(e^{-x})dy=0$
Now check if it is conservative,
$\frac{\partial (-ye^{-x}-3)}{\partial y}=\frac{\partial (e^{-x})}{\partial x}\rightarrow -e^{-x}=-e^{-x}$---Good.
We have turned this differencial equation into an exact differenciatial equation.
Therefore, there exists a function,
$f(x,y)=K$ such as,
$d(f(x,y))=0$ gives our equation above.
---
Let $f(x,y)$ be such a function. Then,
$d(f(x,y))=f_x(x,y)dx+f_y(x,y)dy$
Thus,
$f_x(x,y)=-ye^{-x}-3$
$f_y(x,y)=e^{-x}$
Solving the second partial differential equation yields,
$f(x,y)=ye^{-x}+\phi(x)$ (a function of x alone).
Thus,
$f_x(x,y)=-ye^{-x}+\phi'(x)$
Thus,
$\phi'(x)=-3$
Thus,
$\phi(x)=-3x+C$
Therefore, the function is,
$f(x,y)=-ye^{-x}-3x+C$
Thus,
$-ye^{-x}-3x+C=K$
Thus,
$y=3xe^x+Ce^x$

7. 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?

8. Originally Posted by action259
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?
Affirmative.

$\frac{dy}{dx}=\frac{1-2x}{1+y}$
Which is seperable.

9. Hello, action259!

Check the original wording of the problem, please.
There is no "value of the family of curves".

Find the family of curves if the slope of the tangent at any point is: $\frac{1-2x}{1+y}$

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

If i solve the equation i get: $x - x^2\:=\:y+y^2$ . . . um, not quite
How does this help me?

Separate the variables: . $(1 + y)\,dy \:=\1 - 2x)\,dx" alt="(1 + y)\,dy \:=\1 - 2x)\,dx" />

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

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

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

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

The family is the set of all circles with center at $\left(\frac{1}{2},-\frac{1}{4}\right)$
. . and radius $\sqrt{C + \frac{5}{16}}$ . . . where $C \geq -\frac{5}{16}$