15. Construction of a Second Solution
Dated: 28-11-2024
General case
Consider the differential equation
\[a_{2}(x)\frac{d^{2}y}{dx^{2}}+a_{1}(x)\frac{dy}{dx}+a_{0}(x)y=0\]
Divide the above equation by \(a_2(x)\) to convert this into the form
\[y^{\prime\prime}+P(x)y'+Q(x)y=0\]
Where \(P(x)\) and \(Q(x)\) are continuous on the interval \(I\).
Suppose that \(y_1(x) \ne 0, \forall x \in I\) is a solution to the above differential equation then
\[y_1^{\prime\prime}+P(x)y_1'+Q(x)y_1=0\]
We define \(y = u(x) y_1(x)\) then
\[y^\prime = uy^\prime_1 + y_1u^\prime\]
\[y^{\prime\prime} = uy^{\prime\prime}_1 + 2y^{\prime}_1u^{\prime} + y_1u^{\prime\prime}\]
\[y^{\prime\prime} + Py^\prime + Qy = u(y_1^{\prime\prime} + Py_1^\prime + Qy_1) + y_1u^{\prime\prime} + (2 y_1^\prime + Py_1)u^\prime = 0\]
\[\because y_1^{\prime\prime}+P(x)y_1'+Q(x)y_1=0\]
\[\therefore y_1u^{\prime\prime} + (2 y_1^\prime + Py_1)u^\prime = 0\]
Suppose \(w = u^\prime\), then
\[y_1w^\prime + (2y_1^\prime + Py_1)w = 0\]
Separating the variables, we have
\[\frac {dw}{w} + \left(2 \frac {y_1^\prime}{y_1} + P\right)dx = 0\]
Integrating it
\[ \ln|w| + 2\ln|y_1| = -\int Pdx + c \]
\[ \ln|w(y_1)^2| = -\int Pdx + c \]
\[ w(y_1)^2 = c_1e^{-\int Pdx} \]
\[ w = \frac{c_1e^{-\int Pdx}}{(y_1)^2} \]
\[\because w = u^\prime\]
\[ u^\prime = \frac{c_1e^{-\int Pdx}}{(y_1)^2} \]
Therefore, we integrate again.
\[u = c_1\int \frac{e^{-\int Pdx}}{(y_1)^2}dx + c_2\]
\[\therefore y = u(x)y_{1}(x) = c_{1}y_{1}(x)\int\frac{e^{-\int Pdx}}{y_{1}^{2}}dx + c_{2}y_{1}(x)\]
Choosing \(c_1 = 1\) and \(c_2 = 0\), we obtain
\[y_{2}=y_{1}(x)\int\frac{e^{-\int Pdx}}{(y_{1})^{2}}dx\]
The Woolskin
\[W(y_1(x), y_2(x)) =
\begin{vmatrix}
y_1 & y_1 \int{\frac{e^{- \int Pdx}}{(y_1)^2}dx} \\
(y_1)^\prime & \frac{e^{-\int Pdx}}{y_{1}}+y_{1}^{\prime}\int\frac{e^{-\int Pdx}}{{y_{1}}^{2}}dx
\end{vmatrix}
\]
\[= e^{- \int P dx} \ne 0, \forall x\]
Therefore, \(y_1(x)\) and \(y_2(x)\) are linearly independent and the general solution is
\[y(x) = c_1y_1(x) + c_2y_2(x)\]
Example
Given that
\[y_1 = x^2\]
is a solution of
\[x^2y^{\prime\prime} - 3xy^{\prime} + 4y = 0\]
Find the general solution of the differential equation on the interval \((0, \infty)\).
Solution
The equation can be written as
\[y''-\frac{3}{x}y'+\frac{4}{x^{2}}y=0\]
The second solution \(y_2\) is given by
\[y_{2}=y_{1}(x)\int\frac{e^{-\int Pdx}}{y_{1}^{2}}dx\]
\[y_{2}=x^{2}\int\frac{e^{\int\frac{3}{x}dx}}{x^{4}}dx=x^{2}\int\frac{e^{\ln x^{3}}}{x^{4}}dx\]
\[y_{2}=x^{2}\int\frac{1}{x}dx=x^{2}\ln x\]
Hence the general solution is
\[y = c_1x^2 + c_2x^2 \ln (x)\]
Order Reduction
Given that
\[y_1 = x^3\]
is the solution of differential equation
\[x^2y^{\prime\prime} - 6y = 0\]
Find the second solution of the equation
Solution
Rewrite the equation into
\[y^{\prime\prime}-\frac{6}{x^{2}}y=0\]
So that
\[P(x)=-\frac{6}{x^{2}}\]
\[y_{2}=y_{1}\int\frac{e^{-\int Pdx}}{y_{1}^{2}}dx\]
\[y_{2}=x^{3}\int\frac{e^{-\int \frac{6}{x^{2}}dx}}{x^{6}}dx\]
\[y_{2}=x^{3}\int\frac{e^{\frac{6}{x}}}{x^{6}}dx\]
Therefore, using the formula
\[y_{2}=y_{1}\int\frac{e^{-\int Pdx}}{(y_{1})^{2}}dx\]
This integral is hard to find therefore, we use an alternative approach.
Let
\[y = u(x)y_1(x)\]
\[y = u(x)x^3\]
Then
\[y^{\prime}=3x^{2}u+x^{3}u^{\prime}\]
\[y^{\prime\prime}=x^3u^{\prime\prime}+6x^2u^\prime+6xu\]
Therefore
\[x^{2}y^{\prime\prime}-6y=0\]
\[x^2(x^3u^{\prime}+6x^2u^{\prime}+6xu)-6ux^3=0\]
\[x^5u^{\prime\prime}+6x^4u^\prime=0\]
\[u^{\prime\prime}+\frac{6}{x}u^\prime=0\]
Then if we take \(w = u^\prime\)
\[w^\prime + \frac 6 x w = 0\]
Finding the integrating factor
\[u(x)=e^{\int \frac{1}{x}dx}=e^{6\ln x}=x^{6}\]
Therefore
\[x^6w'+6x^5w=0\]
\[\frac{d}{dx}(x^6w)=0\]
After integration, we have
\[x^6w=c_1\]
\[u'=\frac{c_1}{x^6}\]
Integrating again
\[u=-\frac{c_{1}}{5x^{5}}+c_{2}\]
\[y=ux^{3}=\frac{-c_{1}}{5x^{2}}+c_{2}x^{3}\]
Choosing \(c_2 = 0\) and \(c_1 = -5\), we obtain the second solution
\[y_2 = \frac 1 {x^2}\]
Hence the general solution is
\[y=c_{1}x^{3}+c_{2}\left(\frac{1}{x^{2}}\right)\]
References
Read more about notations and symbols.