Obtain a nonzero solution of ##y''-4y'+x^2(y'-4y)=0## by inspection

[zenterix]

Homework Statement

Obtain a nonzero solution of the equation $y''-4y'+x^2(y'-4y)=0$ by inspection.

This is a problem from Apostol's Calculus and there is no answer for it in the book.

Relevant Equations

$y''-4y'+x^2(y'-4y)=0$

I have not been able to solve this. Here is what I tried to do.

$z=y'-4y$

$z'=y''-4y'$

Thus, the second order equation in $y$ becomes $z'+x^2z=0$, a first order equation in $z$, the solution to which is $z(x)=ke^{-\frac{x^3}{3}}$ with $k>0$.

Thus, $z=ke^{-\frac{x^3}{3}}=y'-4y$.

We can use an integrating factor to solve $y'-4y=ke^{-\frac{x^3}{3}}$ and we arrive at $y(x)=ke^{4x}\int e^{-\frac{t^3}{3}-4t}dt+Ce^{4x}$.

I'm kind of stuck here. I can't solve the integral.

Despite that, we can differentiate this expression and sub in $y'$ and $y$ into the original DE and check that this $y$ is indeed a solution.

Is there a way to get a solution without that integral?

How does one solve the original DE merely "by inspection"?

 

[Orodruin]

I suppose you intended to write $z = y’ - 4y$.

$z=0$ is an obvious solution to the following equation resulting in …

Edit: Note that you are asked for a solution, not all solutions.

 

[zenterix]

If $y'-4y=0$ then $y''-4y'=0$ and we have a solution to the DE.

Then all we have to do is solve $y'-4y=0$ which gives us $y(x)=ke^{4x}$ with $k>0$.

 

[Orodruin]

Exactly. I’d say that is pretty much by inspection.

 

[zenterix]

Exactly. I’d say that is pretty much by inspection.

Indeed it is by inspection.

Now I have another issue. My OP is about a very specific equation. But this problem has basically three parts

1) Let $u$ be a nonzero solution of the 2nd order equation $y''+P(x)y'+Q(x)y=0$. Show that the substitution $y=uv$ converts the equation $y''+P(x)y'+Q(x)y=R(x)$ into a first-order linear equation for $v'$.

2) The question in the original post.

3) Obtain a nonzero solution of the equation $y''-4y'+x^2(y'-4y)=0$ by inspection and use the method of part (a) to find a solution of $y''-4y'+x^2(y'-4y)=2xe^{-x^3/3}$ such that $y(0)=0$ and $y'(0)=4$.

In part 1), I showed that $y''+P(x)y'+Q(x)y=R(x)$ becomes $uv''+v'(2u'+Pu)=R$ when $y=uv$ and $u$ is a solution to $y''+P(x)y'+Q(x)y=0$.

For parts 2 and 3 we have the equation $y''-4y'+x^2(y'-4y)=2xe^{-x^3/3}$ and in my OP I asked about finding a solution to the associated homogeneous equation $y''-4y'+x^2(y'-4y)=0$. This solution is what we take as $u$, namely $u=ke^{4x}$.

Using the calculations from part 1), we can assume that $y(x)=vke^{4x}$ and when we sub this into the nonhomogeneous equation we end up with $v''+4v'=2xe^{-x^3/3}$ which is a simpler equation since it has constant coefficients.

Now, again, I am unsure how to solve this.

What I tried to do just now was to use a theorem that allows us to obtain a solution to the nonhomog. equation by using a linear combination of the solutions to the homogeneous solution, where the weights are determined by very specific integrals. These integrals ended up being things like $\int (-1)\frac{2}{x}e^{-4x+\frac{x^3}{3}}dx$, which I can't solve either.

 

[pasmith]

Indeed it is by inspection.

Now I have another issue. My OP is about a very specific equation. But this problem has basically three parts

1) Let $u$ be a nonzero solution of the 2nd order equation $y''+P(x)y'+Q(x)y=0$. Show that the substitution $y=uv$ converts the equation $y''+P(x)y'+Q(x)y=R(x)$ into a first-order linear equation for $v'$.

2) The question in the original post.

3) Obtain a nonzero solution of the equation $y''-4y'+x^2(y'-4y)=0$ by inspection and use the method of part (a) to find a solution of $y''-4y'+x^2(y'-4y)=2xe^{-x^3/3}$ such that $y(0)=0$ and $y'(0)=4$.

In part 1), I showed that $y''+P(x)y'+Q(x)y=R(x)$ becomes $uv''+v'(2u'+Pu)=R$ when $y=uv$ and $u$ is a solution to $y''+P(x)y'+Q(x)y=0$.

For parts 2 and 3 we have the equation $y''-4y'+x^2(y'-4y)=2xe^{-x^3/3}$ and in my OP I asked about finding a solution to the associated homogeneous equation $y''-4y'+x^2(y'-4y)=0$. This solution is what we take as $u$, namely $u=ke^{4x}$.

Do not include an arbitrary constant here; those will come later, when you add the complementary function (should you need to.)

Using the calculations from part 1), we can assume that $y(x)=vke^{4x}$ and when we sub this into the nonhomogeneous equation we end up with $v''+4v'=2xe^{-x^3/3}$ which is a simpler equation since it has constant coefficients.

Setting $y = e^{4x}v$ I get $$\begin{split} y' - 4y &= e^{4x}(4v + v' - 4v) = v'e^{4x} \\ y'' - 4y' &= e^{4x}(4v' + v'') \end{split}$$ so that $$y'' - 4y' + x^2(y' - 4y) = 2xe^{-x^3/3} \Rightarrow v'' + (4 + x^2)v' = 2xe^{-4x-x^3/3}.$$ This is not consant coefficient, but it is simpler in that it can be solved by an integrating factor.

What I tried to do just now was to use a theorem that allows us to obtain a solution to the nonhomog. equation by using a linear combination of the solutions to the homogeneous solution, where the weights are determined by very specific integrals. These integrals ended up being things like $\int (-1)\frac{2}{x}e^{-4x+\frac{x^3}{3}}dx$, which I can't solve either.

Do not use indefinite integrals here; always specify a lower limit. In this context, changing the lower limit is equivalent to adding a complementary function to the particular solution: the result is still a particular solution.

If the particular solution involves an integral which cannot be done analytically, then it's easiest if the lower limit is the point at which an initial condition is enforced so that the particular solution is known to vanish there, as we always have $$\int_a^a f(x)\,dx = 0.$$ If necessary, a complementary function can then be added to satisfy a non-zero initial condition.

 

[zenterix]

When you reach $v'(x)e^{4x+x^3/3}-v'(0)=x^2$ you still have to integrate to reach $v(x)$ and then obtain $y(x)$ right?

Thus, it seems you reached the same impasse as I did, which is the integral $\int x^2 e^{-4x-x^3/3}dx$.

In my case it was the integrals $t_1(x)=-\int \frac{2}{x} e^{-4x+x^3/3}dx$ and $t_2(x)=\int \frac{2}{x} e^{-8x+x^3/3}dx$ which give the weights for a linear combination of the homogeneous solutions $e^{-4x}$ and $1$, ie

$$y_p(x)=t_1(x)e^{-4x}+t_2(x)$$

Am I missing something or did your calculations above not advance more than my original attempt?

 

[pasmith]

When you reach $v'(x)e^{4x+x^3/3}-v'(0)=x^2$ you still have to integrate to reach $v(x)$ and then obtain $y(x)$ right?

What do you get if you do that? Does the integral become any easier once you substitute the given initial conditions $y(0) = 0$ and $y'(0) = 4$ so that $v(0) = 0$ and $v'(0) = y'(0) - 4y(0) = 4$?

It is fine to leave integrals which cannot be done analytically as definite integrals. In these types of exercises however, the functions are usually chosen such that the integrals can be done; if you end up with something intractable then either you have not gone far enough (for example by leaving constants of integration as arbitrary constants rather than applying the given intial conditions) or have made an error somewhere. In this case you did the latter, when you obtained $v'' + 4v' = 2xe^{-x^3/3}$ rather than the correct $v'' + (4 + x^2)v' = 2xe^{-x^3/3-4x}$.