y” – y = cos hx
This article aims to solve the differential equation. There are two main methods to solve the differential equation: the general solution of homogenous equation and the particular solution of non-homogeneous equation.
Expert Answer
The given differential equation is:
y” – y = cos hxLet’s find a solution to the related homogeneous equation $ y ^{”} -y = 0 $. The auxiliary equation is $ m ^{2} -1=0$ and its solutions are $ m _{1} =-1$ and $ m_ {2} = 1$. The complementary solution is:
\[ y _{c} (x) = c_{ 1 } e ^ { -x } + c_{2} e ^ { x } \]
The function $y _{1} (x) = c_{1} e^{-x}$ and $y_{2} (x) = c_{2} e^{x}$ form a fundamental set.
\[W( y_{1}, y_{2}) = \begin{vmatrix}
y_{1} & y_{2} \\
y’_{1}& y’_{2}
\end{vmatrix}= \begin{vmatrix}
e ^{-x} & e^{x} \\
e^{-x} & e^{x}
\end{vmatrix} = 1+1 = 2\]
Let’s assume that a particular solution has the form $ y_{p} = u_{1}y_{1} + u_{2}y_{2}$ where $ u_{1}$ and $u_{2}$ are the solutions of the systems of equations.
\[y_{1}u’_{1}+ y_{2} u’_{2} = 0\]
\[y_{1}u’_{1} + y_{2} u’_{2} = f (x) \]
$f(x) = \cos hx$ is the input function, using Cramer rule we have:
\[u’_{1} = – \dfrac {y_{2} f(x) }{W}\]
and
\[u’_{2}= \dfrac {y_{1} f(x) }{W}\]
Where $W$ is Wronskian.
Therefore,
\[u’_{1}(x) = -\dfrac { e^{x} .\cos hx }{2} = – \dfrac {1}{4} e^{2x} – \dfrac {1}{4} \]
\[u’_{2}(x) = -\dfrac { e^{-x} .\cos hx }{2} = \dfrac {1}{4} + \dfrac {1}{4} e^{-2x}\]
After integrating both equations,
\[u_{1}(x) = -\dfrac {1}{8} e^{2x} -\dfrac{1}{4}x \]
\[u_{2}(x) = \dfrac {1}{4} x -\dfrac{1}{8} e^{-2x} \]
The particular solution is $y_{p} = u_{1} y_{1} + u_{2}y_{2} $.
Numerical Result
The particular solution of differential equation $y_{p} = u_{1} y_{1} + u_{2}y_{2}$ where:
\[u_{1}(x) = -\dfrac {1}{8} e^{2x} -\dfrac{1}{4}x \]
\[u_{2}(x) = \dfrac {1}{4} x -\dfrac{1}{8} e^{-2x} \]
Example
Solve the given differential equation.\[y” – 6y’ +9y = \dfrac {1}{x} \]Solution
The given differential equation is:
\[ y” – 6y’ +9y = \dfrac {1}{x} \]
Let’s find a solution to the characteristic equation $ r^{2} – 6r+ 9 =0 $. The solutions of the equation is $ r = 3$. The general solution is:
\[y = Ae^{3x} + Bx e^{3x}\]
The fundamental solution and its derivative is:
\[y_{1} (x) = e^{3x} \]
\[y’_{1} (x) = 3e^{3x} \]
\[y_{2} (x) = xe^{3x} \]
\[y’_{2} (x) = (3x+1) e^{3x} \]
The Wronskian is:
\[W(y_{1}, y_{2}) = y_{1}y’_{2} − y_{2} y’_{1} = e^{6x}\]
The general solution is given by:
\[y_{p}(x) = -y_{1}(x) \int \dfrac{y_{2}(x) f(x)}{W(y_{1},y_{2})}dx + y_{2}(x) \int \dfrac{y_{1}(x) f(x)}{W(y_{1},y_{2})}dx\]
By solving the integral $\int \dfrac{y_{2}(x) f(x)}{W(y_{1},y_{2})}$
\[= \dfrac{1}{3} e^{-3x}\]
So,
\[-y_{1}(x) \int \dfrac{y_{2}(x) f(x)}{W(y_{1},y_{2})} = \dfrac {1}{3} \]
By solving the integral $\int \dfrac{y_{1}(x) f(x)}{W(y_{1},y_{2})}$
\[= \int e^{-3x} x^{-6} dx\]
So,
\[ y_{2}(x) \int \dfrac{y_{1}(x) f(x)}{W(y_{1},y_{2})} = x e^{3x} \int e^{-3x} x^{-6} dx\]
Finally, the solution of the differential equation is:
\[y_{p}(x) = \dfrac {1}{3} +x e^{3x} \int e^{-3x} x^{-6} dx\]
