Solve the given differential equation by variation of parameters.

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.

The given differential equation is:                                                  y” –  y =  cos hx Let’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: $y_{p}(x) = \dfrac {1}{3} +x e^{3x} \int e^{-3x} x^{-6} dx$