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) Separate variables.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 $\displaystyle \frac{1}{y}dy=\frac{1}{x-2}-\frac{1}{x+1}dx$
[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
Method #2Originally Posted by action259
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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,
$\displaystyle y=3xe^x+Ce^x$
Method #3
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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,
$\displaystyle y=3xe^x+Ce^x$
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?
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: $\displaystyle \frac{1-2x}{1+y}$
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 $\displaystyle C \geq -\frac{5}{16}$