6 L16: Reduction Of Order (5.6)
6.1 Reduction of order Technique
Reduction of order is a technique used to find a second, linearly independent solution to a second-order homogeneous differential equation when one solution is already known. This method is particularly useful because knowing two linearly independent solutions to a homogeneous linear differential equation allows us to write the general solution to the equation.
The process essentially involves using the known solution to reduce the order of the differential equation, hence the name “reduction of order.” By doing so, you transform the problem into a first-order differential equation, which is often easier to solve. After solving this reduced equation, you can obtain the second solution, which, combined with the first, provides a complete solution to the original differential equation.
This method is especially useful in situations where the standard methods for solving differential equations (such as characteristic equations for linear equations with constant coefficients) are not applicable or when the equation is too complex for direct methods.
\[p_0(x)y''+p_1(x)y'+p_2(x)y=F(x)\] Suppose \(y_1\) is a solution to
\[p_0(x)y''+p_1(x)y'+p_2(x)y=0\]
Let \[\boxed{y=uy_1}\rightarrow \boxed{y'=u'y_1+uy_1'}\rightarrow \boxed{y''=u''y_1+2u'y_1'+uy_1''}\]
\[\Rightarrow\] \[p_0(u''y_1+2u'y_1'+uy_1'')+p_1(u'y_1+uy_1')+p_2uy_1=F(x)\] \[\Rightarrow\] \[p_0y_1u''+(2p_0y_1'+p_1y_1)u'+{(p_0y_1''+p_1y_1'+p_2y_1)}u=F(x)\] But \(y_1\) is a solution of the homogeneous, so last term of LHS is zero.
\[\Rightarrow \boxed{p_0y_1u''+(2p_0y_1'+p_1y_1)u'=F(x)}\]
Let \[\boxed{Z=u'}\text{ and }\boxed{q_0=p_0y_1}\text{ and } \boxed{q_1=2p_0y_1'+p_1y_1}\]
Which allow us to transform a 1st order:
\[q_0Z'+q_1Z=F(x)\]
To obtain in standard form:
\[\boxed{\boxed{Z'+\frac{q_1}{q_0}Z=\frac{F(x)}{q_0}}}\qquad\text{(1st Order Linear)}\]
6.2 Example 1. LNCCNH
Linear, Non constant coefficients, Non Homogeneous and solution for the homogeneous is given
Find the general solution of
\[xy''-(2x+1)y'+(x+1)y=x^2\] and \(y_1=e^x\) is a solution for the homogeneous.
\[\boxed{y=ue^x}\rightarrow \boxed{y'=u'e^x+ue^x}\rightarrow \boxed{y''=u''e^x+2u'e^x+ue^x}\] \[x(u''e^x+2u'e^x+ue^x)-(2x+1)(u'e^x+ue^x)+(x+1)ue^x=x^2\]
Now let’s factor by \(u\), \(u''\) and \(u'\): \[u({xe^x-(2x+1)e^x+(x+1)e^x})+u''(xe^x)+u'(2xe^x-(2x+1)e^x)=x^2\]
\[u''xe^x-u'(2xe^x-(2x+1)e^x)=x^2\] \[u''xe^x-u'(2xe^x-2xe^x+e^x)=x^2\] \[u''xe^x-u'e^x=x^2\]
Let \(Z=u',\quad q_0=xe^x,\quad q_1=e^x\),
\[\Rightarrow q_0Z'-q_1Z=x^2\] \[\Rightarrow Z'-\frac{q_1}{q_0}Z=\frac{x^2}{q_0}\] \[\Rightarrow Z'-\frac{e^x}{xe^x}Z=\frac{x^2}{xe^x}\] \[\Rightarrow Z'-\frac{1}{x}Z=\frac{x}{e^x}\]
\[\Rightarrow Z'-\frac{1}{x}Z=xe^{-x}\] Now we can use method for First Order Linear:
\[\mu=e^{\int p\, dx}=e^{\int -\frac{1}{x}\,dx}=e^{-\ln(x)}=e^{\ln(x^{-1})}=\frac{1}{x}\] Then
\[Z=\frac{1}{\mu}\int\mu Q\,dx=x\int e^{-x}\,dx=-x(e^{-x}+c_1)\]
Then
\[u=\int Z\,dx=\int (c_1x-xe^{-x})\,dx=c_1\frac{x^2}{2}+(x+1)e^{-x}+c_2\]
Finally:
\[\boxed{y=uy_1=ue^x=c_1\frac{x^2}{2}e^x+(x+1)+c_2e^x}\]
6.3 Example 2. LNCCH
Linear, Non constant coefficients, Homogeneous and a solution is given
Find the general solution, and the fundamental set for
\[x^2y''-3xy'+3y=0\] given the solution \(y_1=x\)
\[\boxed{y=ux}\rightarrow\boxed{y'=u'x+u}\rightarrow\boxed{y''=u''x+u'+u'}\] \[\Rightarrow x^2(u''x+2u')-3x(u'x+u)+3ux=0\] \[u{(3x-3x)}+u''(x^3)+u'(2x^2-3x^2)=0\] \[u''(x^3)+u'(2x^2-3x^2)=0\] \[x^3u''-x^2u'=0\] \[Z=u'\Rightarrow q_0=x^3,\quad q_1=x^2\] \[Z'-\frac{1}{x}Z=0\qquad\text{(Separable)}\]
\[\int \frac{1}{Z}dZ=\int\frac{1}{x}\,dx\] \[\ln Z = \ln x + c_1\] \[\ln Z = \ln x + \ln(c_2)\] \[\ln Z = \ln c_2x\] \[Z=c_2x\]
\[\Rightarrow u=\int Z\,dx=\int c_2x\,dx=c_2\frac{x^2}{2}+c_3\]
Finally:
\[\boxed{y=uy_1=ux=c_2\frac{x^3}{2}+c_3x}\] Therefore the fundamental set of solution for the diffyq is :\[\{x,x^3\}\]
6.4 Summary
- Find a particular solution for the homogeneous: \(y_1\).
- Assume \(y=uy_1\) is a solution.
- Use the solution at the diffyQ.
- Factor by \(u\), then see which factor is zero(satisfy homogeneous).
- Change of variables: \(Z=u'\), and the factors in front of \(Z'\) and \(Z\), call them \(q_0\) and \(q_1\) respectively.
- State the standard form for
one-order-lessdiffyQ. - Solve the one-order-less-diffyQ by any method which is convenient.
- Integrate \(Z\) to find \(u\).
- State your general solution as \(y=uy_1\).
- Check you have the same numbers of constants as the order of your original diffyQ.