Prove a substitution converts a given second order equation to a first order equation

by RoRi

February 23, 2016

Consider the second-order differential equation
$y'' + P(x) y' + Q(x) y = 0$

and let $u$ be a solution to the equation. Show that the substitution $y = uv$ converts the equation

$y'' + P(x)y' + Q(x)y = R(x)$

into a first-order liner equation for $v'$ .
By inspection, find a nonzero solution of the second order differential equation
$y'' - 4y' + x^2 (y' - 4y) = 0$

and use part (a) to find a solution of the differential equation

$y'' - 4y' + x^2(y' - 4y) = 2xe^{-\frac{x^3}{3}}$

with

$y = 0 \quad \text{and} \quad y' = 4 \quad \text{when } x = 0.$

Incomplete.

7 comments

April 22, 2019 at 11:24 am

Evangelos says:

Every solution comment I post disappears. WordPress go to hell.

Reply
April 22, 2019 at 11:23 am

Evangelos says:

a.) If we make the substitution $y = uv$ , with $u$ being a nonzero function satisfying

$\begin{align*} y'' + P(x)y' + Q(x)y &= 0 \end{align*}$

The equation

$\begin{align*} y'' + P(x)y' + Q(x)y &= R(x) \end{align*}$

Becomes

$\begin{align*} (u''v + 2u'v' + v''u) + P(x)(u'v + v'u) + Q(x)(uv) \end{align*}$

Grouping terms by multiples of v and its derivatives gives us

$\begin{align*} v''u + v'(2u' + P(x)u) + v(u'' + P(x)u' + Q(x)u) &= R(x) \\ \end{align*}$

Since u satisfies $y'' + P(x)y' + Q(x)y = 0$ , we get

$\begin{align*} v''u + v'(2u' + P(x)u) &= R(x) \\ \end{align*}$

And since $u$ is a nonzero function, we can divide both sides by it, giving us

$\begin{align*} v'' + v'(2\frac{u'}{u} + P(x)) &= \frac{R(x)}{u} \end{align*}$

If we let $F(x) = (2\frac{u'}{u} + P(x))$ and $G(x) = \frac{R(x)}{u}$ , the above equation becomes

$\begin{align*} v'' + F(x)v' &= G(x) \end{align*}$

Which is a linear first-order equation of v’, as requested.

Reply
April 22, 2019 at 11:21 am

Evangelos says:

a.) If we make the substitution $y = uv$ , with $u$ being a nonzero function satisfying

$\begin{align*} y'' + P(x)y' + Q(x)y &= 0 \end{align*}$

The equation

$\begin{align*} y'' + P(x)y' + Q(x)y &= R(x) \end{align*}$

Becomes

$\begin{align*} (u''v + 2u'v' + v''u) + P(x)(u'v + v'u) + Qx)(uv) \end{align*}$

Grouping terms by multiples of v and its derivatives gives us

$\begin{align*} v''u + v'(2u' + P(x)u) + v(u'' + P(x)u' + Q(x)u) &= R(x) \\ \end{align*}$

Since u satisfies $y'' + P(x)y' + Q(x)y = 0$ , we get

$\begin{align*} v''u + v'(2u' + P(x)u) &= R(x) \\ \end{align*}$

And since $u$ is a nonzero function, we can divide both sides by it, giving us

$\begin{align*} v'' + v'(2\frac{u'}{u} + P(x)) &= \frac{R(x)}{u} \end{align*}$

If we let $F(x) = (2\frac{u'}{u} + P(x))$ and $G(x) = \frac{R(x)}{u}$ , the above equation becomes

$\begin{align*} v'' + F(x)v' &= G(x) \end{align*}$

Which is a linear first-order equation of v’, as requested.

b.) By inspection, $u = e^{4x}$ is a nonzero function that satisfies

$\begin{align*} y'' - 4y' + x^{2}(y' - 4y) &= 0 \end{align*}$

Now, if we use the method of substitution from part (a.), the equation

$\begin{align*} y'' - 4y' + x^{2}(y' - 4y) &= 2xe^{-x^{3}/3} \end{align*}$

Becomes

$\begin{align*} v''(u) + v'(2u' + (x^{2} - 4)) &= 2xe^{-x^{3}/3} \end{align*}$

And since we know that $u = e^{4x}$ , the above equation becomes

$\begin{align*} v'' + v(x^{2} + 4) &= 2xe^{-(\frac{x^{3}}{3} + 4x)} \end{align*}$

Which is a linear equation of $v'$ . If we let

$\begin{align*} P(x) &= x^{2} + 4 \\ Q(x) &= 2xe^{-(\frac{x^{3}}{3} + 4x)} \\ A(x) &= \int_{0}^{x} t^{2} + 4 \quad dt \\ &= \frac{x^{3}}{3} + 4x \end{align*}$

Then

$\begin{align*} v' &= e^{-A(x)}(v'(0) + \int_{0}^{x}Q(t)e^{A(t)} dt) \\ &= e^{-(\frac{x^{3}}{3} + 4x)}(v'(0) + x^2) \end{align*}$

Now, we can deduce the value of v'(0) using our given information. We know that $y' = u'v + v'u$ , and we also know that when $x = 0$ , $u = u' = 1$ , and $y = uv = 0$ , so $v$ must be 0. Plugging in those values gives us $y' = v' = 4$ when $x = 0$ . Thus:

$\begin{align*} v' &= e^{-(\frac{x^{3}}{3} + 4x)}(x^{2} + 4) \end{align*}$

But if that is the case, then we can integrate $v'$ to give us

$\begin{align*} v &= \int_{0}^{x} e^{-(\frac{t^{3}}{3} + 4t)}(t^{2} + 4) dt \\ &= 1 -e^{-(\frac{x^{3}}{3} + 4x)} \end{align*}$

And since $y = uv$ , we can multiply to get

$\begin{align*} y &= e^{4x} - e^{-x^{3}/3} \end{align*}$

Reply
April 21, 2019 at 10:52 pm

Evangelos says:

Testing inline latex formatting $x = 0$ should be x = 0

Reply
- April 21, 2019 at 10:54 pm
  
  Evangelos says:
  
  I typed out an entire reply and it disappeared D:
  
  Reply
  - April 22, 2019 at 10:43 am
    
    Evangelos says:
    
    Testing both inline and separate line $x = 0$
    
    $\begin{align*} y &= mx + b \end{align*}$
    
    I’m still salty my post got lost
April 21, 2019 at 10:49 pm

Evangelos says:

a.) We start with our substitution of y = uv

$\begin{align*} y &= uv \\ y' &= u'v + v'u \\ y'' &= u''v + 2u'v' + v''u\\ \end{align*}$

And the equation

$\begin{align*} y'' + P(x)y' + Q(x)y &= R(x) \end{align*}$

Becomes

$\begin{align*} (u''v + 2u'v' + v''u) + P(x)(u'v + v'u) + Q(x)(uv) &= R(x) \end{align*}$

Now, if we group together terms by v and its derivatives, we get

$\begin{align*} v''u + v'(2u' + P(x)u) + v(u'' + P(x)u' + Q(x)u) &= R(x) \end{align*}$

But, we know from our given information that u satisfies

$\begin{align*} y'' + P(x)y' + Q(x)y &= 0 \end{align*}$

So the factor multiplying v becomes 0, giving us

$\begin{align*} v''u + v'(2u' + P(x)u) &= R(x) \end{align*}$

Since u is a nonzero function, we can divide both sides by it, giving us

$\begin{align*} v'' + v'(2\frac{u'}{u} + P(x)) &=\frac{R(x)}{u} \end{align*}$

Now, if we define two new functions of x as follows

$\begin{align*} F(x) &= 2\frac{u'}{u} + P(x) \\ G(x) &= \frac{R(x)}{u} \end{align*}$

The above differential equation becomes

$\begin{align*} v'' + F(x)v' &= G(x) \end{align*}$

Which is a first-order linear equation of v’

b.) By inspection, we are to find a nonzero function u = f(x) that satisfies the equation

$\begin{align*} y'' -4y' + x^{2}(y' - 4y) &= 0 \end{align*}$

The following function satisfies the equation

$\begin{align*} u &= e^{4x} \end{align*}$

Now, we are to use the function u and the information from part (a.) to find a solution to the following equation

$\begin{align*} y'' -4y' + x^{2}(y' - 4y) &= 2xe^{-x^{3}/3} \end{align*}$

From part (a.), we know that we can use the substitution $y = uv$ to turn the above equation into a first-order linear equation of $v'$ , as follows

$\begin{align*} (u''v + 2u'v' + v''u) - 4(u'v + v'u) + x^{2}((u'v + v'u) - 4(uv)) &= 2xe^{-x^{3}/3} \end{align*}$

$\begin{align*} v''u + v'(2u' - 4u + x^{2}u) + v(u'' -4u' + x^{2}((u' - 4u)) &= 2xe^{-x^{3}/3} \end{align*}$

And of course, from our inspection, we know that $u'' -4u' + x^{2}((u' - 4u) = 0$ , so

$\begin{align*} v''u + v'(2u' - 4u + x^{2}u) &= 2xe^{-x^{3}/3} \end{align*}$

And we know that $u = e^{4x}$ , so the above equation becomes

$\begin{align*} v''(e^{4x}) + v'(4e^{4x} + x^{2}e^{4x}) &= 2xe^{-x^{3}/3} \end{align*}$

And since the exponential function is never 0 for any real x, we can divide both sides by $e^{4x}$ to give

$\begin{align*} v'' + v'(4 + x^{2}) &= 2xe^{-(\frac{x^3}{3} + 4x)} \end{align*}$

Now, since this is a linear equation of v’, we can use the equation from Theorem 8.3, along with our given values when $x = 0$ , to give an equation for v’. Let

$\begin{align*} P(x) &= (x^{2} + 4)\\ Q(x) &= 2xe^{-(\frac{x^3}{3} + 4x)} A(x) &= \int_{0}^{x} (t^{2} + 4) dt \\ &= \frac{x^{3}}{3} + 4x \end{align*}$

Our equation for $v'$ becomes

$\begin{align*} v' &= e^{-A(x)}(v'(0) + \int_{0}^{x}Q(t)e^{A(t)} dt \\ &= e^{-(\frac{x^{3}}{3} + 4x)}(v'(0) + x^{2}) \end{align*}$

Going back to our given information, we know that $y = uv$ or $v = \frac{y}{u}$ , so when $x = 0$ , $v = 0$ since $y = 0$ and $u = 1$ . As such, $y'(0) = u'(0)v(0) + v'(0)u(0) = v'(0) = 4$ . Giving us

$\begin{align*} v' &= e^{-(\frac{x^{3}}{3} + 4x)}(4 + x^{2}) \end{align*}$

Integrating this gives us

$\begin{align*} v &= -e^{-(\frac{x^{3}}{3} + 4x)} + C \end{align*}$

And since we know that $v(0) = 0$ , the constant C must be 1. Now, to get y, we go back to our original substitution $y = uv$

$\begin{align*} y &= uv \\ &= (e^{4x})(-e^{-(\frac{x^{3}}{3} + 4x)} + 1) \\ &= e^{4x} - e^{-x^{3}/3} \end{align*}$

Reply

Stumbling Robot

Prove a substitution converts a given second order equation to a first order equation

Related

7 comments

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply