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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.

Expert Answer

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: \[y_{p}(x) = \dfrac {1}{3} +x e^{3x} \int e^{-3x} x^{-6} dx\]

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