Homogenous Differential Equation Solution

[_N3WTON_]

Homework Statement

Solve the differential equation:
$\frac{dy}{dx} = x + y$

Homework Equations

Integrating factor equation

The Attempt at a Solution

Ok, I am aware that in order to solve this equation, I need to make a substitution:
$v = x + y$
However, at this point I am unsure about what to do. Unfortunately, my DiffEq textbook does not cover this technique thoroughly. I am hoping that perhaps somebody can explain how this technique works, thanks.

 

[Ray Vickson]

Homework Statement

Solve the differential equation:
$\frac{dy}{dx} = x + y$

Homework Equations

Integrating factor equation

The Attempt at a Solution

Ok, I am aware that in order to solve this equation, I need to make a substitution:
$v = x + y$
However, at this point I am unsure about what to do. Unfortunately, my DiffEq textbook does not cover this technique thoroughly. I am hoping that perhaps somebody can explain how this technique works, thanks.

The substitution $v = x+y$ leads to another inhomogeneous DE, but simpler than the original. Have you tried it?

I cannot believe that your textbook does not cover the methods for solving
$$\frac{dy}{dx} + f(x) y = g(x)$$
You say it does not cover this technique thoroughly, but surely it covers it at least a little bit. You need to show your work, and explain where/why you are stuck.

 

[LCKurtz]

@N3WTON: Alternatively, doesn't your text discuss constant coefficient homogeneous and non-homogeneous equations and the method of undetermined coefficients for NH equations?

 

[ehild]

Homework Statement

Solve the differential equation:
$\frac{dy}{dx} = x + y$

The Attempt at a Solution

Ok, I am aware that in order to solve this equation, I need to make a substitution:
$v = x + y$
However, at this point I am unsure about what to do. Unfortunately, my DiffEq textbook does not cover this technique thoroughly. I am hoping that perhaps somebody can explain how this technique works, thanks.

With the substitution, y=v-x, y'=v'-1 and the differential equation becomes v'-1=v, that is, v'=v+1. Is it not a separable equation?

ehild

 

[_N3WTON_]

@N3WTON: Alternatively, doesn't your text discuss constant coefficient homogeneous and non-homogeneous equations and the method of undetermined coefficients for NH equations?

It does discuss constant coefficient homogenous and non-homogenous equations, however; no mention of the method of undetermined coefficients for NH equations. Perhaps we have just not reached that point in the text yet

 

[_N3WTON_]

The substitution $v = x+y$ leads to another inhomogeneous DE, but simpler than the original. Have you tried it?

I cannot believe that your textbook does not cover the methods for solving
$$\frac{dy}{dx} + f(x) y = g(x)$$
You say it does not cover this technique thoroughly, but surely it covers it at least a little bit. You need to show your work, and explain where/why you are stuck.

I have tried it but I don't understand what I need to do. The equation you listed could be solved using the integrating factor method, no? I'm confused about how that relates to the equation I listed...

 

[_N3WTON_]

With the substitution, y=v-x, y'=v'-1 and the differential equation becomes v'-1=v, that is, v'=v+1. Is it not a separable equation?

ehild

ok...so picking up from there I did:
$\int\frac{v'}{v+1} = \int 0$
$ln(v+1) = C$
$v + 1 = e^c$
$v = e^{c} - 1$
$x + y = e^{c} - 1$
Here I am stuck because the solution in the book is:
$y = 2e^x - x - 1$
Also the book notes that this problem can be "solved quite easily" using the integrating factor method, but I do not see it...

 

[pasmith]

ok...so picking up from there I did:
$\int\frac{v'}{v+1} = \int 0$

This should be $$\int \frac{v'}{v +1}\,dx = \int 1\,dx.$$

$ln(v+1) = C$
$v + 1 = e^c$
$v = e^{c} - 1$
$x + y = e^{c} - 1$
Here I am stuck because the solution in the book is:
$y = 2e^x - x - 1$

You appear to have omitted an initial condition (perhaps y(0) = 1) from your problem; otherwise '2' should instead be an arbitrary constant.

 

[ehild]

ok...so picking up from there I did:
$\int\frac{v'}{v+1} = \int 0$

What is v+1 divided by (v+1)??
ehild

 

[_N3WTON_]

This should be $$\int \frac{v'}{v +1}\,dx = \int 1\,dx.$$

nvm

 

[_N3WTON_]

thanks, but can you explain why that is correct?
before I had:
$v' = v+1$
[itex]\frac{v'}{v+1} = 0 [itex]
I am not seeing where the dx comes from...

 

[Orodruin]

∫v′v+1,dv=∫0

You seem to have assumed that $v'/(v+1) = 0$. This is not the case so I suggest you recheck that algebra.

Another option is to find a particular solution (there is a fairly obvious one) and then add the solution to the homogeneous equation to that (which works since the ODE is linear).

 

[_N3WTON_]

This should be $$\int \frac{v'}{v +1}\,dx = \int 1\,dx.$$
You appear to have omitted an initial condition (perhaps y(0) = 1) from your problem; otherwise '2' should instead be an arbitrary constant.

you're right, I do apologize for that...I'll make the edit now

 

[_N3WTON_]

What is v+1 divided by (v+1)??
ehild

ok, I see the mistake now...thank you

 

[pasmith]

thanks, but can you explain why that is correct?
before I had:
$v' = v+1$
$\frac{v'}{v+1} = 0$
I am not seeing where the dx comes from...

You are integrating both sides with respect to $x$. It is crucial to state what variable you are integrating with respect to, because shortly you will use substitution to change $\int \frac{v'}{v +1}\,dx$ into an integral with respect to $v$.

 

[_N3WTON_]

You are integrating both sides with respect to $x$. It is crucial to state what variable you are integrating with respect to, because shortly you will use substitution to change $\int \frac{v'}{v +1}\,dx$ into an integral with respect to $v$.

thank you, so here is my second attempt:
$\int\frac{v'}{v+1} dx = \int dx$
$ln(v+1) = x + C$
$v+1 = e^x + C$
$x + y = e^x + C$

 

[ehild]

thank you, so here is my second attempt:
$\int\frac{v'}{v+1} dx = \int dx$
$ln(v+1) = x + C$
$v+1 = e^x + C$

the last line is wrong. It should be $v+1 = e^{x + C}$ instead.

 

[_N3WTON_]

the last line is wrong. It should be $v+1 = e^{x + C}$ instead.

thank you...

 

[Ray Vickson]

I have tried it but I don't understand what I need to do. The equation you listed could be solved using the integrating factor method, no? I'm confused about how that relates to the equation I listed...

In your example, is it not the case that $f(x) = -1$ and $g(x) = x$?

 

[ehild]

the last line is wrong. It should be $v+1 = e^{x + C}$ instead.

And $v+1 = e^{x + C} = Ae^x$ . The constants C or A depend on the initial condition.

ehild

 

[_N3WTON_]

And $v+1 = e^{x + C} = Ae^x$ . The constants C or A depend on the initial condition.

ehild

ok, the initial condition was f(0) = 1...you arrived at that answer just by bringing the constant C in front of the exponential?

 

[_N3WTON_]

In your example, is it not the case that $f(x) = -1$ and $g(x) = x$?

ok so if i were to use the integrating factor method I would set:
$p(x) = -1$ ?
I'm still a bit confused about how I could do this problem using integrating factor method...I understand how to use separation of variables..

 

[Ray Vickson]

ok so if i were to use the integrating factor method I would set:
$p(x) = -1$ ?
I'm still a bit confused about how I could do this problem using integrating factor method...I understand how to use separation of variables..

There is a standard formula; just apply it to your problem.

 

[ehild]

ok, the initial condition was f(0) = 1...you arrived at that answer just by bringing the constant C in front of the exponential?

You know that $e^{{x+C}}=e^C e^x$, but e^C is also a constant, I denoted it by A. So v+1=y+x+1=Ae^x.. Applying the condition y(0)=1, 1+1=A.

 

[Orodruin]

Just for completeness, since we have a linear ODE with constant coefficients and a polynomial inhomogeneity, we could simply have assumed a particular solution of the form $y_p = ax + b$ and arrived at $a = b = -1$ and thus $y_p = -x-1$, $y'_p = -1$. We then write $y = y_p + y_h$ where $y_h$ satisfies the homogeneous ODE $y'_h = y_h$, which I believe anyone familiar with differential equations can solve ... This approach is what I was trying to hint at in post #12.

 

[HallsofIvy]

It is better NOT to memorize formulas (like that for the integrating factor of a linear differential equation) but to understand the basic concept and work from there. Here, the differential equation is dy/dx= x+ y which is the same as dy/dx- y= x. An "integrating factor" for this equation is a a function u(x) such that multiplying by it makes the left side a "complete derivative". That is, so that udy/dx- uy= d(uy)/dx. Using the product rule on the right, u dy/dx- uy= u dy/dx+ u'y. We can cancel the "u dy/dx" terms leaving -uy= u'y or u'= -u. Integrating that gives $u(x)= Ce^{-x}$.

Since we only need one, take C= 1 to get $u(x)= e^{-x}$. Multiplying both sides of the equation by that, we have $e^{-x}dy/dx- e^{-x}y= d(e^{-x}y)/dx= xe^{-x}$. Integrate both sides of that with respect to x.

 

[_N3WTON_]

You know that $e^{{x+C}}=e^C e^x$, but e^C is also a constant, I denoted it by A. So v+1=y+x+1=Ae^x.. Applying the condition y(0)=1, 1+1=A.

Thank you...your insight has been very helpful to me :)

 

[_N3WTON_]

Just for completeness, since we have a linear ODE with constant coefficients and a polynomial inhomogeneity, we could simply have assumed a particular solution of the form $y_p = ax + b$ and arrived at $a = b = -1$ and thus $y_p = -x-1$, $y'_p = -1$. We then write $y = y_p + y_h$ where $y_h$ satisfies the homogeneous ODE $y'_h = y_h$, which I believe anyone familiar with differential equations can solve ... This approach is what I was trying to hint at in post #12.

I appreciate your help, and I understand your solution. However, that is not a technique that we have covered in my DiffEq class so I don't know if I'd be able to use it on my upcoming exam...

 

[_N3WTON_]

It is better NOT to memorize formulas (like that for the integrating factor of a linear differential equation) but to understand the basic concept and work from there. Here, the differential equation is dy/dx= x+ y which is the same as dy/dx- y= x. An "integrating factor" for this equation is a a function u(x) such that multiplying by it makes the left side a "complete derivative". That is, so that udy/dx- uy= d(uy)/dx. Using the product rule on the right, u dy/dx- uy= u dy/dx+ u'y. We can cancel the "u dy/dx" terms leaving -uy= u'y or u'= -u. Integrating that gives $u(x)= Ce^{-x}$.

Since we only need one, take C= 1 to get $u(x)= e^{-x}$. Multiplying both sides of the equation by that, we have $e^{-x}dy/dx- e^{-x}y= d(e^{-x}y)/dx= xe^{-x}$. Integrate both sides of that with respect to x.

thank you, that actually helped me to better understand what I'm actually doing when I apply the integrating factor method...

 

[Orodruin]

I appreciate your help, and I understand your solution. However, that is not a technique that we have covered in my DiffEq class so I don't know if I'd be able to use it on my upcoming exam...

I think that the approach itself is not very different from the variable substitution that you started with (just writing $v = y + x + 1$ rather than $v = y+x$ - you could also see it as making yet another substitution from $v' = v + 1$ to $u = v+1$, which also gives $u' = u$). Regardless, only a very ignorant teacher would deduct points for you using a working method unless explicitly asking for something to be solved with a particular method. But of course this is completely up to you.

 

[_N3WTON_]

I think that the approach itself is not very different from the variable substitution that you started with (just writing $v = y + x + 1$ rather than $v = y+x$ - you could also see it as making yet another substitution from $v' = v + 1$ to $u = v+1$, which also gives $u' = u$). Regardless, only a very ignorant teacher would deduct points for you using a working method unless explicitly asking for something to be solved with a particular method. But of course this is completely up to you.

thats true...to my knowledge the test doesn't explicitly ask for a certain method so thanks for help :)

 

[ehild]

Thank you...your insight has been very helpful to me :)

You are welcome:)

ehild