Solve the differential equation

In summary: Just need to solve for c2 now.In summary, the conversation involved solving a differential equation and finding a new one with no constants. This was done by using basic algebra and integration techniques, as well as understanding the concepts of integrating factors for linear equations and using initial conditions to solve for constants.
  • #1
hbomb
58
0
I need someone to look over my work if possible

1) Solve the differential equation
xdy=(5y+x+1)dx

Here is what I did:
x=(5y+x+1)dx/dy
[tex]x=5xy+\frac{x^2}{2}+x[/tex]
[tex]0=5xy+\frac{x^2}{2}[/tex]
[tex]\frac{-x^2}{2}=5xy[/tex]
[tex]y=\frac{-x}{10}[/tex]

2) Solve: [tex]y(x^2-1)dx+x(x^2+1)dy=0[/tex]

Here is what I did:

[tex]x^2ydx-ydx+x^3dy+xdy=0[/tex]

[tex]\frac{xdy-ydx}{x^2}+ydx+xdy=0[/tex]

[tex]d(\frac{y}{x})+d(xy)=0[/tex]
This is where I am stuck at. I need to have something multiply the first differential that will yield something in the form of [tex]\frac{y}{x}[/tex] and also have the same thing multiply the second differential and will yield something in the form of xy.

3) Find a differential equation with the solution [tex]y=c_1sin(2x+c_2)[/tex]

Here is what I did:

[tex]y'=2c_1cos(2x+c_2)[/tex]

and since
[tex]c_1=\frac{y}{sin(2x+c_2)}[/tex]

[tex]y'=\frac{2ycos(2x+c_2)}{sin(2x+c_2}[/tex]

[tex]y'=2ycot(2x+c_2)[/tex]

I'm not sure what to do after this. I know I need to somehow get rid of the [tex]c_2[/tex] constant, but I don't know how to do this.
 
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  • #2
I'm in a rush, so answers are quite incomplete and tend to be misleading, here they go
1) it seems suitable for y=vx, dy=xdv/dx + v substitution

2) [tex]y(x^2-1)dx+x(x^2+1)dy=0[/tex]

[tex]y(1-x^2)dx = x(x^2+1)dy[/tex]
[tex](x^2-1)dx/x(x^2+1) = dy/y[/tex]
simply integrate both sides.

3) how about giving initial conditions? ;)
 
  • #3
Answer for 3)

Hi! I think I have the answer for 3).
Here it goes:

y=c1sin(2x+2) (E.1)

y=c1sin(2x+2) ---> y^2=[c1sin(2x+2)]^2 (E.2)

y=c1sin(2x+2) ---> y'/2=c1cos(2x+2) ---> (y'/2)^2=[c1cos(2x+2)]^2 (E.3)

By adding (E.2) and (E.3) you have:

y^2 + (y'/2)^2=(c1)^2 because (cosk)^2 + (sink)^2 = 1

From (E.1) c1=y/sin(2x+2)

We finally write:

y^2 + (y'/2)^2=[y/sin(2x+2)]^2
 
  • #4
for equation 3 shouldn't it be y'/2=4c1cos(2x+c2) sin(2x+c2) for the middle step, because you have to use the chain rule.
 
  • #5
First, since this was clearly labled "homework help", I've moved it to the homework section!

hbomb said:
I need someone to look over my work if possible

1) Solve the differential equation
xdy=(5y+x+1)dx

Here is what I did:
x=(5y+x+1)dx/dy
[tex]x=5xy+\frac{x^2}{2}+x[/tex]
This makes no sense at all. You appear to have integrated the right hand side with respect to x, treating y as a consant, even though is clearly a function of x, but have not done anything to left hand side. Didn't you learn in algebra "whatever you do to one side of an equation, you must do to the other"?
[tex]0=5xy+\frac{x^2}{2}[/tex]
[tex]\frac{-x^2}{2}=5xy[/tex]
[tex]y=\frac{-x}{10}[/tex]

If this were true, then dy/dx= -1/10. Putting that and y= -x/10 into the original equation gives -x/10= -5x/10+ x+ 1. No, that is not true for all x.

Rewrite the original equation as xdy/dx- 5y= 1- x and then divide by x:
dy/dx- (5/x)y= 1/x- x. That's a linear equation. Do you know how to find an integrating factor for a linear differential equation?

2) Solve: [tex]y(x^2-1)dx+x(x^2+1)dy=0[/tex]

Here is what I did:

[tex]x^2ydx-ydx+x^3dy+xdy=0[/tex]
[tex]\frac{xdy-ydx}{x^2}+ydx+xdy=0[/tex]

I had to look at this closely: Factoring x2 out of x2ydx and x3dy, you get x2(ydx+ xdy) That leaves xdy- ydx so you divide the entire equation by x2 to get that.

[tex]d(\frac{y}{x})+d(xy)=0[/tex]
This is where I am stuck at. I need to have something multiply the first differential that will yield something in the form of [tex]\frac{y}{x}[/tex] and also have the same thing multiply the second differential and will yield something in the form of xy.[\quote]
??I don't see why you are stuck. Now, just integrate that equation:
[tex]\frac{y}{x}+ xy= C[/tex]
Or are you saying you don't see HOW to get that equation? You don't need to multiply by anything:
[tex]d\frac{y}{x}= \frac{xdy- ydx}{x^2}[/tex]
and
[tex]d(xy)= ydx+ xdy[/tex]

3) Find a differential equation with the solution [tex]y=c_1sin(2x+c_2)[/tex]

Here is what I did:

[tex]y'=2c_1cos(2x+c_2)[/tex]

and since
[tex]c_1=\frac{y}{sin(2x+c_2)}[/tex]

[tex]y'=\frac{2ycos(2x+c_2)}{sin(2x+c_2}[/tex]

[tex]y'=2ycot(2x+c_2)[/tex]

I'm not sure what to do after this. I know I need to somehow get rid of the [tex]c_2[/tex] constant, but I don't know how to do this.
Since you have two constants, the differential equation must be second order. Yes, y'= 2c1cos(2x+ c2) so y"= -4c1sin(2x+c2= -4y. That was easy!
 
  • #6
Thanks for the help. For question 3 I'm suppose to have a new equation with no c1 or c2 in it. In other words I'm suppose to solve for c1, then take the derivative of the original equation, substitute c1 in for that, then take the derivative of the new equation, then solve for c2. Here is an easy problem to demonstrate what I mean.

[tex]y=c_1e^{2x}+c_2e^{-x}[/tex]

[tex]y'=2c_1e^{2x}-c_2e^{-x}[/tex]

[tex]y"=4c_1e^{2x}+c_2e^{-x}[/tex]

[tex]y'+y=3c_1e^{2x}[/tex]

[tex]y"+y'=6c_1e^{2x}[/tex]

[tex]c_1=\frac{y'+y}{3e^{2x}}[/tex]

[tex]y"+y'=2y'+2y[/tex]

[tex]y"-y'-2y=0[/tex]

See...no c1's or c2's
 
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1. What is a differential equation?

A differential equation is an equation that describes the relationship between a function and its derivatives. It involves the use of derivatives to solve for the unknown function.

2. Why is solving differential equations important?

Solving differential equations is important in many fields of science and engineering, as it allows us to model and predict the behavior of complex systems. It is also the basis for many mathematical models used in physics, chemistry, economics, and more.

3. How do you solve a differential equation?

The approach to solving a differential equation depends on its type and complexity. Some methods include separation of variables, using specific formulas, or using numerical techniques. It is important to understand the properties and characteristics of the equation before choosing a method.

4. What are the applications of differential equations?

Differential equations are used to model a wide range of real-world phenomena such as population growth, heat transfer, fluid dynamics, and electrical circuits. They are also used in fields like economics, biology, and medicine to make predictions and analyze data.

5. Can differential equations be solved analytically?

Some differential equations can be solved analytically, meaning a closed-form solution can be found. However, many equations do not have closed-form solutions and require numerical techniques for approximation. Additionally, some equations may not have any solution at all.

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