Index
Let \( (m,n) \in \hspace{0.04em} \mathbb{N}^2 \) be two natural numbers and:
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a series of continuous functions \( a_1(x), a_2(x), \ ... \, a_n(x) \)
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a series of functions \( f_1(x), f_2(x), \ ... \, f_m(x) \)
Let \( y(x) \) be a function of class \( \mathbb{C}^{n}\) on an interval \(I\). We note \(y^{(n)}\) its \(n\)-th derivative .
In the context of solving a linear differential equation of order \(n\) where the right hand side is a linear combination of functions such that \( (E)\):
with for all \( k \in [\![ 1, m ]\!] \), the function \( (y_k) \) as a specific solution of the equation \( (E_k) \) :
The superposition principle tells us that:
Proofs
Let \( (m,n) \in \hspace{0.04em} \mathbb{N}^2 \) be two natural numbers and:
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- a series of continuous functions \( a_1(x), a_2(x), \ ... \, a_n(x) \)
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- a series of functions \( f_1(x), f_2(x), \ ... \, f_m(x) \)
Let \( y(x) \) be a function of class \( \mathbb{C}^{n}\) on an interval \(I\). We note \(y^{(n)}\) its \(n\)-th derivative .
We start from the equation \( (E) \), linear differential of order \( n\) where the right hand side is a linear combination of functions.
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Specific solution for each function \( f_k(x)\)
For all \( k \in [\![ 1, m ]\!] \), we then have a series of equations \( (\tilde E_k) \) to solve:
$$ \left \{ \begin{gather*} a_n(x) y^{(n)}(x) + \ ... \ + \hspace{0.2em} a_1(x) y'(x) + a_0(x) y(x) = f_1(x) \qquad (\tilde E_1) \\ a_n(x) y^{(n)}(x) + \ ... \ + \hspace{0.2em} a_1(x) y'(x) + a_0(x) y(x) = f_2(x) \qquad (\tilde E_2) \\ ........................ \\ a_n(x) y^{(n)}(x) + \ ... \ + \hspace{0.2em} a_1(x) y'(x) + a_0(x) y(x) = f_k(x) \qquad (\tilde E_k) \\ ........................ \\ a_n(x) y^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) y'(x) + a_0(x) y(x) = f_m(x) \qquad (\tilde E_m) \\ \end{gather*} \right \}$$In this series of equations, we then notice that:
$$ \forall k \in [\![ 1, m ]\!], \enspace y_k \enspace solution \enspace of \enspace (\tilde E_k) $$If each \( y_k \) is a solution for \( (\tilde E_k) \), then each \( y_k \) verifies:
$$ a_n(x) y_k^{(n)}(x) + \ ... \ + \hspace{0.2em} a_1(x) y_k'(x) + a_0(x) y_k(x) = f_k(x) \qquad (\tilde E_k)$$Thereby, by multiplying \( (\tilde E_k) \) by \( \lambda_k \):
$$ a_n(x) \lambda_k y_k^{(n)}(x) + \ ... \ + \hspace{0.2em} a_1(x) \lambda_k y_k'(x) + a_0(x) \lambda_k y_k(x) = \lambda_k f_k(x) \qquad ( \lambda_k \tilde E_k)$$Now, thanks to linearity of the derivative , we know that:
$$ \bigl( \lambda f \bigr)' = \lambda f' $$So in our case,
$$ \forall k \in [\![ 1, m]\!], \enspace \bigl( \lambda_k y_k \bigr)' = \lambda_k y'_k \qquad (1) $$Thanks to\( (1) \), we can rearrange it and see that:
$$ a_n(x) (\lambda_k y_k)^{(n)}(x) + \ ... \ + a_1(x) \bigl(\lambda_k y_k \bigr)'(x) + a_0(x) \bigl(\lambda_k y_k\bigr)(x) = \lambda_k f_k(x) \qquad ( \lambda_k \tilde E_k)$$This equation shows that:
$$ \forall k \in [\![ 1, m]\!], \enspace \lambda_k y_k \enspace solution \enspace of \enspace (\lambda_k \tilde E_k) $$ -
Total specific solution from the aggregation of functions \( \lambda_k f_k \)
Previously, we were able to see that each \( \lambda_k y_k \) is solution for \( (\lambda_k \tilde E_k) \):
$$ \left \{ \begin{gather*} a_n(x) (\lambda_1 y_1)^{(n)}(x) + \ ... \ + a_1(x) \bigl(\lambda_1 y_1 \bigr)(x)' + a_0(x) \bigl(\lambda_1 y_1\bigr)(x) = \lambda_1 f_1(x) \qquad (\lambda_1 \tilde E_1) \\ a_n(x) (\lambda_2 y_2)^{(n)}(x) + \ ... \ + a_1(x) \bigl(\lambda_2 y_2\bigr)(x) ' + a_0(x) \bigl(\lambda_2 y_2\bigr)(x) = \lambda_2 f_2(x) \qquad (\lambda_2 \tilde E_2) \\ ........................ \\ a_n(x) (\lambda_k y_k)^{(n)}(x) \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + a_1(x) \bigl(\lambda_k y_k \bigr)(x)' + a_0(x) \bigl(\lambda_k y_k \bigr)(x) = \lambda_k f_k(x) \qquad (\lambda_k \tilde E_k) \\ ........................ \\ a_n(x) (\lambda_m y_m)^{(n)}(x) + \ ... \ + a_1(x) \bigl(\lambda_m y_m\bigr)(x) ' + a_0(x) \bigl(\lambda_m y_m \bigr)(x) = \lambda_m f_m(x) \qquad (\lambda_m \tilde E_m) \\ \end{gather*} \right \} $$By adding up the \( (\lambda_k \tilde E_k) \) from \(1 \) to \(m \):
$$ a_n(x) \underbrace{\Biggl[ \sum_{k=1}^m \lambda_k y_k^{(n)} \Biggr]} _{ y_s^{(n)} } \hspace{0.2em} + \hspace{0.2em} ... \hspace{0.2em} + \hspace{0.2em} a_1(x) \underbrace{\Biggl[ \sum_{k=1}^m \lambda_k y_k' \Biggr]} _{ y_s' } \hspace{0.2em} + \hspace{0.2em} a_0(x) \underbrace{\Biggl[ \sum_{k=1}^m \lambda_k y_k \Biggr]} _{ y_s } \hspace{0.2em} = \hspace{0.2em} \sum_{k=1}^m \lambda_k f_k \qquad (E^*) $$Thus, a total specific solution \( y_s \) which will be the addition of all the specific solutions \( \lambda_k y_k \):
$$ y_s(x) = \lambda_1 y_1(x) + \lambda_2 y_2(x) + \ ... \ + \hspace{0.2em} \lambda_m y_m(x) $$In the end, thanks to \( (E^*) \), we see that this function is indeed a solution of \( ( E) \):
$$ a_n(x) y_s^{(n)}(x) + \ ... \ + a_1(x) y_s'(x) + a_0(x) y_s(x) = \lambda_1 f_1(x) + \lambda_2 f_2(x) + \ ... \ + \hspace{0.2em} \lambda_m f_m(x) \qquad (E) $$And as a result,
$$ \forall k \in [\![ 1, m ]\!], $$$$ y_k \enspace \underline{\text{ specific solution of }} (\tilde E_k) \Longleftrightarrow (\lambda_1 y_1 + \lambda_2 y_2 \hspace{0.2em} + \ ... \ + \hspace{0.2em} \lambda_m y_m) \underline{\text{ total specific solution of }} (E) $$
Examples
Let \( (E) \) be a first order linear differential equation \( LDE_1 \) with a constant coefficient \( (H) \) its associated homogeneous equation:
We then have a series of equations \( (\tilde E_1), (\tilde E_2), (\tilde E_3) \) to solve:
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Solving the first equation \( (\tilde E_1)\)
This equation was solved in the example of solving \( LDE_1 \) with constant coefficient .
The specific solution \( y_1 \) of \( (\tilde E_1) \) is:
$$ y_1(x)= \frac{x^2 }{2}-\frac{x}{2} + \frac{1}{4} $$ -
Solving the second equation \( (\tilde E_2)\)
We have a solution to the homogeneous equation \( (H) \) (see the example of solving \( EDL_1 \) with constant coefficient ):
$$ y_h(x) = Ke^{-2x} $$We then look for a particular solution \( y_s \) of type:
$$ y_2(x) = K(x) e^{-2x} \qquad (y_2) $$After having performed a variation of parameters , we seek to determine the function \( K(x) \) such as:
$$ K'(x) = \cos(x)e^{2x} $$Now, taking its antiderivative , we do have:
$$ K(x) = \int^x \cos(t)e^{2t} \hspace{0.2em} dt$$We perform an integration by parts with the following choice for \( u \) and \( v' \):
$$ \Biggl \{ \begin{gather*} u(t) = \cos(t) \\ v'(t) = e^{2t } \end{gather*} $$$$ \Biggl \{ \begin{gather*} u'(t) = -\sin(t)\\ v(t) = \frac{1}{2} e^{2t } \end{gather*} $$$$K(x) = \frac{1}{2} \Biggl[\cos(t) e^{2t }\Biggr]^x - \int^x \frac{1}{2} (-\sin(t)) \hspace{0.2em} e^{2t } \hspace{0.2em} dt $$$$K(x) = \frac{1}{2} \Biggl[\cos(t)e^{2t }\Biggr]^x + \int^x \frac{1}{2} \sin(t) \hspace{0.2em} e^{2t } \hspace{0.2em} dt $$$$K(x) = \frac{1}{2} \Biggl[\cos(t) e^{2t }\Biggr]^x + \frac{1}{2} \Biggl( \frac{1}{2} \Biggl[\sin(t) e^{2t }\Biggr]^x - \frac{1}{2}\int^x \cos(t) e^{2t} \hspace{0.2em} dt \Biggr) $$We note that \( K(x) \) reappears, so we replace it:
$$K(x) = \frac{1}{2} \Biggl[\cos(t) e^{2t }\Biggr]^x + \frac{1}{4} \Biggl[\sin(t) e^{2t }\Biggr]^x - \frac{1}{4} K(x) $$$$ \frac{5}{4} K(x) = \frac{1}{2} \cos(x)e^{2x } + \frac{1}{4} \sin(x) e^{2x } $$$$ K(x) =\frac{8}{20} \cos(x) e^{2x } + \frac{4}{20} \sin(x) e^{2x } $$$$ K(x) = e^{2x } \biggl( \frac{2}{5}\cos(x) + \frac{1}{5} \sin(x) \biggr) \qquad (K) $$Injecting \( K \) into \( y_s \), the exponentials annihilate:
$$ y_2(x) = e^{2x } \biggl( \frac{2}{5} \cos(x) + \frac{1}{5} \sin(x) \biggr) e^{-2x} $$The specific solution \( y_2 \) of \( (\tilde E_2) \) is:
$$ y_2(x)= \frac{2 }{5}\cos(x) +\frac{1}{5}\sin(x) $$ -
Solving the third equation \( (\tilde E_3)\)
$$ y'(x) + 2 y(x) = 1 \qquad \qquad (\tilde E_3) $$Here, \(\frac{1 }{2} \) is an obvious solution.
Thus, the specific solution \( y_3 \) of \( (\tilde E_3) \) is:
$$ y_3(x)= \frac{1 }{2} $$ -
Superposition of solutions\(: \sum \lambda_k y_k \)
We saw in the demonstration above that:
$$ \forall k \in [\![ 1, m ]\!], $$$$ y_k \ \underline{\text{ specific solution of }} (\tilde E_k) \Longleftrightarrow (\lambda_1 y_1 + \lambda_2 y_2 \hspace{0.2em} + \ ... \ + \hspace{0.2em} \lambda_m y_m) \underline{\text{ total specific solution of }} (E) $$So in our case:
$$ 2 y_1(x) + 3 y_2(x) + y_3(x) \ \underline{total \ specific \ solution} \ of \ ( E) $$Let us then calculate this particular total solution \(y_s\):
$$ y_s(x) = 2 y_1(x) + 3 y_2(x) + y_3(x) $$$$ y_s(x) = 2 \Biggl( \frac{x^2 }{2}-\frac{x}{2} + \frac{1}{4} \Biggr) + 3 \Biggl(\frac{2 }{5}\cos(x) +\frac{1}{5}\sin(x) \Biggr) + \frac{1}{2} $$$$ y_s(x) = x^2 - x + \frac{1}{2} + \frac{6 }{5}\cos(x) +\frac{3}{5}\sin(x) + \frac{1}{2} $$$$ y_s(x) = x^2 - x + 1 + \frac{6 }{5}\cos(x) +\frac{3}{5}\sin(x) $$ -
Verification of the total specific solution \( y_s\)
If \( y_s \) is solution for \( (E) \), then:
$$ y_s'(x) + 2 y_s(x) = 2x^2 + 3\cos(x) + 1 $$Let us check it.
$$ y_s'(x) + 2 y_s(x) = \Biggl( 2x -1 -\frac{6 }{5} \sin(x) +\frac{3}{5} \cos(x)\Biggr) + 2\Biggl( x^2 - x + 1 + \frac{6 }{5}\cos(x) +\frac{3}{5}\sin(x) \Biggr) $$$$ y_s'(x) + 2 y_s(x) = 2x -1 -\frac{6 }{5} \sin(x) +\frac{3}{5} \cos(x) + 2x^2 - 2x +2 + \frac{12 }{5}\cos(x) + \frac{6}{5}\sin(x)$$By tidying up a little:
$$ y_s'(x) + 2 y_s(x) = 2x^2 + \hspace{0.2em} \underbrace{ 2x - 2x} _{ =0 } \hspace{0.2em} + \hspace{0.2em} 2 -1 + \hspace{0.2em} \underbrace{\frac{6}{5}\sin(x) -\frac{6 }{5} \sin(x)} _{ =0 } \hspace{0.2em} + \frac{12 }{5}\cos(x) +\frac{3}{5} \cos(x) $$$$ y_s'(x) + 2 y_s(x) = 2x^2 +1 + \frac{15}{5}\cos(x) $$$$ y_s'(x) + 2 y_s(x) = 2x^2 + 3\cos(x) +1 $$We definitely verified that \( y_s \) was a solution for \( (E) \).
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