Kansas State University

Mathematics Department

Math 340 Home, Textbook Contents, Online Homework Home

Warning: MathJax requires JavaScript to process the mathematics on this page.
If your browser supports JavaScript, be sure it is enabled.

Change of Variables

Discussion

Basically, there is only one way to solve a first order differential equation. That is to convert it to exact form and integrate it. We have applied this to exact equations, which are already in exact form; to separable equations, which are in exact form after they are separated; and to linear equations, which are in exact form after they are multiplied by an integrating factor. There is one other standard approach to putting a first order equation in exact form, making a change of variables. Two instances where this works are Bernoulli equations and homogeneous equations. (WARNING: The term homogeneous has several different meanings in differential equations. We will encounter the term later with a completely different meaning in Chapter 2.)

Bernoulli Equations

An equation is a Bernoulli equation if it can be written in the form $$ \frac{dy}{dx} + p(x)y = q(x)y^n $$ for some $n$. Bernoulli equations are almost linear equations, they just have an extra $y^n$ term. We can make a change of variables to get rid of this term and rewrite the equation as a linear equation. Let $y = v^{1/(1-n)}$. Then $y^n=v^{n/(1-n)}$ and $$ \frac{dy}{dx} = \frac{1}{1-n}v^{1/(1-n)-1}\frac{dv}{dx} =\frac{1}{1-n}v^{n/(1-n)}\frac{dv}{dx} $$ so plugging into our equation we obtain $$ \frac{1}{1-n}v^{n/(1-n)}\frac{dv}{dx} + p(x)v^{1/(1-n)}=q(x)v^{n/(1-n)} $$ Dividing through by $[1/(1-n)]v^{n/(1-n)}$ we obtain $$ \frac{dv}{dx} + (1-n)p(x)v = (1-n)q(x) $$ which is linear. We now solve the linear equation using ordinary techniques. Finally, since the original problem is stated in terms of $x$ and $y$, the answer should be given in terms of $x$ and $y$ --- not $v$. So we undo our substitution by writing $y=v^{1/(1-n)}$. We must also check for singular solutions of the form $y=0$ since the substitution of $y=v^{1/(1-n)}$ is equivalent to dividing by $y$ if $n>1$.

Paradigm

$\displaystyle \frac{dy}{dx}+3y=e^xy^2. $ STEP 1: Make the substitution $y = v^{1/(1-n)}$.

Here $n=2$ so $y = v^{-1}$ $$ \begin{align} \frac{dy}{dx}&=-v^{-2}\frac{dv}{dx} \\ y^2&=v^{-2} \end{align} $$ and the equation is $$ -v^{-2}\frac{dv}{dx}+3v^{-1}=e^xv^{-2}. $$ STEP 2: Divide through to obtain a linear equation $$ \frac{dv}{dx}-3v=-e^x. $$ STEP 3:Solve the linear equation for v

We follow the paradigm for a linear equation.

  1. $\mu(x) = e^{\int -3\,dx} = e^{-3x}$

  2. $e^{-3x}\frac{dv}{dx}-3e^{-3x}v=-e^{-2x}$

  3. $\frac{d}{dx}(e^{-3x}v)=-e^{-2x}$

  4. $e^{-3x}v=\int -e^{-2x}\,dx=(1/2)e^{-2x}+C$

  5. $v=(1/2)e^x+Ce^{3x}$

STEP 4: Back substitute to find $y$ $$ y=v^{-1}=[(1/2)e^x+Ce^{3x}]^{-1} $$ STEP 5: Check for singular solution $y=0$. $$0+3\cdot0=e^x\cdot0$$ so $y=0$ is a solution as well.

Example

$$\frac{dy}{dx}+y=\cos(x)/y, \qquad y(0)=1$$ FIRST: Find the general solution.

Step 1: Let $y=v^{1/2}$ then $$(1/2)v^{-1/2}\frac{dv}{dx}+v^{1/2}=\cos(x)/v^{1/2}$$ Step 2: $\displaystyle\frac{dv}{dx}+2v=2\cos(x)$

Step 3: Solving for $v$

  1. $\mu(x)=e^{\int2\,dx}=e^{2x}$

  2. $\displaystyle e^{2x}\frac{dv}{dx}+2e^{2x}v=2e^{2x}\cos(x)$

  3. $\displaystyle \frac{d}{dx}(e^{2x}v)=2e^{2x}\cos(x)$

  4. $\displaystyle e^{2x}v=\int 2e^{2x}\cos(x)\,dx = e^{2x}\left(\frac45\cos(x)+\frac25\sin(x)\right)+C $

  5. $\displaystyle v(x)=\left(\frac45\cos(x)+\frac25\sin(x)\right)+Ce^{-2x}$

Step 4: $\displaystyle y(x)=\sqrt{\frac45\cos(x)+\frac25\sin(x)+Ce^{-2x}} $

Step 5: $y=0$ is not a solution (can't have $\cos(x)/0$ on right-hand-side).

SECOND: Plug in the initial value and solve for the arbitrary constant. $$ \begin{align} y(0)=\sqrt{\frac45\cos(0)+\frac25\sin(0)+Ce^0}&{\buildrel \text{set} \over = } \sqrt{\frac45+C}&=1 \\ C&=\frac15 \end{align} $$ So $\displaystyle y(x)=\sqrt{\frac45\cos(x)+\frac25\sin(x)+\frac15e^{-2x}} $

You can generate additional examples of initial value problems for first order Bernoulli equations here.

Homogeneous Equations

An equation is homogeneous if it can be written in the form $$ \frac{dy}{dx}=f(y/x) $$ for some function $f$. Usually, it takes some algebraic manipulation to convert the equation to this form. Often, the equation is given in the form $$ \frac{dy}{dx}=\frac{a_nx^n+a_{n-1}x^{n-1}y+a_{n-2}x^{n-2}y^2+\ldots+a_0y^n}{ b_nx^n+b_{n-1}x^{n-1}y+b_{n-2}x^{n-2}y^2+\ldots+b_0y^n} $$ In this case, divide through by $x^n$ to obtain the desired form. Once the equation is in the desired form, we make the change of variables $v=y/x$ so that $y=xv$ and $dy/dx=xdv/dx+v$. Plugging into our equation we obtain $$ x\frac{dv}{dx}+v=f(v) $$ which is separable. We now solve the separable equation using ordinary techniques. Finally, since the original problem is stated in terms of $x$ and $y$, the answer should be given in terms of $x$ and $y$, not $v$. So we undo our substitution by writing $y = xv$.

Paradigm

$$ \frac{dy}{dx}=\frac{2xy}{x^2+y^2} $$ Step 0: Convert to homogeneous form

In this case, every term has order 2 so we divide by $x^2$ to obtain $$ \frac{dy}{dx}=\frac{2(y/x)}{1+(y/x)^2} $$ Step 1: Make the substitution $v=y/x$

This is equivalent to $y=xv$ so we get $dy/dx=x\,dv/dx+v$ and plugging into the equation yields $$ x\frac{dv}{dx}+v=\frac{2v}{1+v^2} $$ Step 2: Solve the separable equation

We use the paradigm for separable equations:

  1. Separate the variables $$ \begin{align} x\frac{dv}{dx}&=\frac{2v}{1+v^2}-v=\frac{v-v^3}{1+v^2} \\ \frac{(1+v^2)dv}{v-v^3}&=\frac{dx}{x} \end{align} $$
  2. Integrate both sides $$ \log(\frac{v}{v^2-1})=\log(x)+C $$
  3. Solve for $v$ $$ \begin{align} \frac{v}{v^2-1}&=kx \\ v&=\frac{1\pm\sqrt{1+4k^2x^2}}{2kx} \end{align} $$
  4. Check for singular solutions.

    We divided by $v^3-v$ which is $0$ at $v=0$ and $v=\pm1$. None of these is included in the general solution so they are all singular solutions.

Step 3: Back Substitute $v=y/x$ to find $y$ $$ y=\frac{1\pm\sqrt{1+4k^2x^2}}{2k} $$ is the general solution and the singular solutions are $$ \begin{align} y&=0 \\ y&=\pm x. \end{align} $$

Example

$$\frac{dy}{dx}=\frac{x+y}{y},\qquad y(1)=1/2+\sqrt5/2$$ FIRST: Find the general solution.

Step 1: Let $v=y/x$ (or $y=xv$) to get $$ x\frac{dv}{dx}+v=\frac{v+1}{v} $$ Step 2: We now solve this equation.

  1. $$ \begin{align} x\frac{dv}{dx}&=\frac{v+1}{v}-v \\ x\frac{dv}{dx}&=\frac{v+1-v^2}{v} \\ \frac{v\,dv}{v+1-v^2}&=\frac{dx}{x} \end{align} $$
  2. $$ \begin{align} \int\frac{v\,dv}{v+1-v^2}&=\int\frac{dx}{x} \\ \left(\frac{\sqrt5}{10}-\frac12\right)\log(2v+\sqrt5-1) - \left(\frac{\sqrt5}{10}+\frac12\right)\log(2v-\sqrt5-1)&=\log(x)+C \\ \frac{(2v+\sqrt5-1)^{\sqrt5/10-1/2}}{(2v-\sqrt5-1)^{\sqrt5/10+1/2}}&=kx \end{align} $$
  3. I can't solve this for $v$ to get an explicit solution.

  4. We divided by $v+1-v^2$ whose roots are $v=1/2\pm\sqrt5/2$. These are both solutions to the equation and only $v=1/2-\sqrt5/2$ is an example of the general solution (corresponding to $k=0$), so $v=1/2+\sqrt5/2$ is a singular solution.

Step 3: Replace $v$ by $y/x$ to get $$ \frac{(2y/x+\sqrt5-1)^{\sqrt5/10-1/2}}{(2y/x-\sqrt5-1)^{\sqrt5/10+1/2}}=kx $$ with a singular solution $y/x=1/2+\sqrt5/2$.

SECOND: Plug in the initial value and solve for the arbitrary constant. $$ \begin{align} \frac {(2\times(1/2+\sqrt5/2)/1+\sqrt5-1)^{\sqrt5/10-1/2}}{ (2\times(1/2+\sqrt5/2)/1-\sqrt5-1)^{\sqrt5/10+1/2}} &=k\times1 \\ \frac{(2\sqrt5)^{\sqrt5/10-1/2}}{0}&=k \end{align} $$ Well that didn't work out very well. There is no way to choose $k$ to get an example of the general solution which satisfies the initial condition. But we have one more solution left to consider. The singular solution is $y/x=1/2+\sqrt5/2$ and it does satisfy $y(1)=1/2+\sqrt5/2$. So the solution to our initial value problem is $$ y=\frac{(1+\sqrt5)x}2 $$ This problem involves somewhat more complicated numbers than most in this text. That has nothing to do with the solution to the initial value problem being the singular solution. Homogeneous equations often give rise to integrals with ugly answers.

You can generate additional examples of initial value problems for first order homogeneous equations here.

If you have any problems with this page, please contact bennett@math.ksu.edu.
©2010, 2014 Andrew G. Bennett