You CAN Ace Differential Equations

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Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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This technique takes a second-order, linear differential equation with one known solution and reduces the equation to a first-order, linear equation that may possibly be solved using a first-order technique. In general, the idea can be applied to even higher order equations but, due to complexity of integration, this technique is really only used on second-order equations. However, it can still be useful in many cases.

classification: second-order, linear, homogeneous

\(y'' + p(t)y' + q(t)y = 0\)

The idea is that we have an equation of the form \(y'' + p(t)y' + q(t)y = 0\) and we know one solution, \(y_1(t)\). To get a second solution, we try \(y_2=v(t)y_1(t)\), i.e. we assume that the second solution is equal to the first solution times some function of t. This may seem unusual but it works in many cases. Now, let's work through the calculus and see where it gets us.

For simplicity, letting \(y=y_2\), we have \(y=v(t)y_1(t)\). Taking the derivative with respect to t, we get \(y'=v'y_1+vy_1'\) and \(y''=v''y_1+2v'y_1'+vy_1''\).
To summarize, we have

\(y=vy_1\)

\(y'=v'y_1+vy_1'\)

\(y''=v''y_1+2v'y_1'+vy_1''\)

Next, we take these expressions and substitute them back into the original differential equation \(y'' + p(t)y' + q(t)y = 0\). After collecting terms, we get
\( y_1v'' + (2y_1'+py_1)v' + (y_1''+py_1'+qy_1)v = 0 \)
Notice that the expression in parentheses of the last term is just the original differential equation, which is zero. So we end up with
\( y_1v'' + (2y_1'+py_1)v' = 0 \)

Now, let \(w=v'\) in the last equation to get \( y_1w' + (2y_1'+py_1)w = 0 \). This is now a first-order, linear equation, which we can use to solve for w. Finally, we integrate w to get v and we are done.

Let's take a minute to watch the next video, which explains why this works.

PatrickJMT - Reduction of Order - Why It Works [13mins-27secs]

video by PatrickJMT

Specific Case - Constant Coefficients with Real, Repeated Roots

On the second-order, linear page, we introduced the idea that, in the case of real, repeated roots, we would have one solution \(y_1=e^{rt}\). To show that the second solution is \(y_2=ty_1\), we will use reduction of order in this example. (Test yourself by trying this on your own before looking at the solution.)

Let \(y_1=c_1e^{rt}\) be one solution to \(ay''+by'+cy=0\). Since r is a repeated root of the characteristic equation, we can write the differential equation as \(y''-2ry'+r^2y=0\). Use reduction of order to derive the second solution \(y_2=c_2te^{rt}\).

For simplicity, we will write \(y_2=v(t)y_1(t)=v(t)c_1e^{rt}\) as \(y=cve^{rt}\).
Our first step is to calculate the derivatives. We will need to use the product rule.

\(y=cve^{rt}\)

\(y'=cvre^{rt}+cv'e^{rt} =\) \(ce^{rt}(vr+v')\)

\(y''=ce^{rt}[v'r+v''] + cre^{rt}[vr+v'] = \) \(ce^{rt}[v'r+v''+vr^2+v'r] =\) \(ce^{rt}[v''+2rv'+r^2v]\)

Now we substitute each of these into the original differential equation \(y''-2ry'+r^2y=0\).
\(ce^{rt}[v''+2rv'+r^2v]- \) \(2rce^{rt}[vr+v']+r^2cve^{rt}=0\)
\(ce^{rt}[v''+2rv'+r^2v-\) \(2r^2v-2rv'+r^2v]=0\)
The \(v\) and \(v'\) terms cancel, leaving only \(ce^{rt}v''=0\). Since \(ce^{rt}\) is never zero, we are left with \(v''=0\).
Now we integrate twice to get
\(v''=0 \to v'=k \to v=kt+p\) where k and p are the constants of integration.

Now we have \(y_2=ce^{rt}[kt+p]\). Next, we combine the two solutions into one. We were not asked to do this in the problem statement but doing this will allow us to explain what to do with all the constants.
\( \begin{array}{rcl} y &=& y_1+y_2 = ce^{rt}+ckte^{rt}+cpe^{rt} \\ &=& e^{rt}(c+cp)+kc(te^{rt}) \\ &=& c_1e^{rt}+c_2te^{rt} \\ &=& e^{rt}[c_1+c_2t] \end{array} \)
In the work above, we let \(c_1=c+cp\) and \(c_2=kc\), in essence just renaming the constants.

final answers

second solution

\(y_2=c_2te^{rt}\)

general solution

\(y=e^{rt}[c_1+c_2t]\)

close solution

Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, find the second solution and the general solution to each differential equation using reduction of order. If initial conditions are given, solve the initial value problem also. Give your answers in exact terms and completely factored.

Determine the second fundamental solution of \( x^2y'' - 6y = 0 \), given \( y_1 = x^3 \)

Problem Statement

Determine the second fundamental solution of \( x^2y'' - 6y = 0 \), given \( y_1 = x^3 \)

Final Answer

\( y_2 = x^{-2} \) is the second fundamental solution.

Problem Statement

Determine the second fundamental solution of \( x^2y'' - 6y = 0 \), given \( y_1 = x^3 \)

Solution

We will use the technique of Reduction of Order, where \( y_2 = vy_1 \) and \(v\) is a function of \(x\).

\( y_2 = vy_1 \)

\( y'_2 = v(3x^2) + x^3 v' \)

\( y''_2 = v(6x) + 3x^2 v + x^3 v'' + 3x^2 v' \)

\( y''_2 = x^3 v'' + 6x^2 v' + 6xv \)

\( x^2 y''_2 - 6y_2 = x^2[ x^3 v'' + 6x^2 v' + 6xv] - 6vx^3 \)

\( x^5 v'' + 6x^4 v' = 0 \)

Let \( u = v' \).
\( \begin{array}{rcl} x^5 u' + 6x^4u & = & 0 \\ x^5 u' & = & -6x^4 u \\ \displaystyle{\frac{u'}{u}} & = & \displaystyle{\frac{-6}{x}} \\ \displaystyle{\frac{du}{u}} & = & \displaystyle{\frac{-6}{x}dx} \\ \ln|u| & = & -6\ln|x| + c_1 \\ u & = & c_1 x^{-6} \\ v' & = & c_1 x^{-6} \\ dv & = & c_1 x^{-6}dx \\ \displaystyle{v = \frac{c_1 x^{-5}}{-5} + k} \end{array}\)

\(\displaystyle{ y_2 = v x^3 = \left[ \frac{c_1 x^{-5}}{-5} + k \right] x^3 = \frac{c_1x^{-2}}{-5} + kx^3 }\)

let \( c = c_1 / (-5) \)

\( y_2 = c/x^2 + kx^3 \)

Since \( kx^3 \) is a multiple of \( y_1 \) it is already included in \( y_1 \), so we just drop that term.

Final Answer

\( y_2 = x^{-2} \) is the second fundamental solution.

close solution

\( x^2y'' + 5xy' - 5y = 0, x > 0; y_1 = x \)

Problem Statement

\( x^2y'' + 5xy' - 5y = 0, x > 0; y_1 = x \)

Solution

1814 solution video

video by PatrickJMT

close solution

\( x^2y'' - 3xy' + 4y = 0; y_1 = x^2 \)

Problem Statement

\( x^2y'' - 3xy' + 4y = 0; y_1 = x^2 \)

Solution

1815 solution video

video by MIP4U

close solution

\( t^2y'' - t(t+2)y' + (t+2)y = 0; y_1 = t \)

Problem Statement

\( t^2y'' - t(t+2)y' + (t+2)y = 0; y_1 = t \)

Solution

1816 solution video

video by Educator.com

close solution

\( x^2y'' + 3xy' + y = 0; y_1 = 1/x \)

Problem Statement

\( x^2y'' + 3xy' + y = 0; y_1 = 1/x \)

Solution

1817 solution video

video by MIP4U

close solution
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