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Practice question solving linear homogeneous second order differentials using the auxiliary equation method.

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Solve the following second order differential

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$\\frac{d^{2}y}{dx^{2}}\\var{a}\\frac{dy}{dx}\\var{b}y=0$ where $y=\\var{c}$ and $\\frac{dy}{dx}=2$ when $x=0$.

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$\\frac{d^{2}y}{dx^{2}}\\var{a}\\frac{dy}{dx}\\var{b}y=0$ where $y=\\var{c}$ and $\\frac{dy}{dx}=2$ when $x=0$.

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Simplify the differential by substituting $D=\\frac{d}{dx}$ and extract $y$

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$(D^{2}\\var{a}D\\var{b})y=0$ 

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Convert the terms between the brackets to an auxiliary equation by replacing $D$ with $m$ to give

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$(m^{2}\\var{a}m\\var{b})y=0$ 

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Find the roots of m usign the quadratic roots formula

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\\[\\frac{-b \\pm \\sqrt{b^2-4ac}}{2a}  \\]

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Where $a=1$, $b=\\var{a}$ and $c=\\var{b}$

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\\[\\frac{-\\var{a} \\pm \\sqrt{\\var{a}^2-4(1)(\\var{b})}}{2}  \\]

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Roots: m = \\[ \\var{Rootalpha} \\] or \\[ \\var{Rootbeta} \\]

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Roots are Real and Different

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General solution will be in the form $y = Ae^{\\alpha x}+Be^{\\beta x}$

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Let the roots of m equal $\\alpha$ and $\\beta$

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$\\alpha = \\var{Rootalpha}$ and $\\beta = \\var{RootBeta}$

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$y = Ae^{\\var{Rootalpha}x}+Be^{\\var{Rootbeta}x}$...general solution

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To find the particular solution, sub in the inital conditions: $y=\\var{c}$ and $\\frac{dy}{dx}=2$ when $x=0$.

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$\\var{c} = Ae^{\\var{Rootalpha}(0)}+Be^{\\var{Rootbeta}(0)}$

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$\\var{c} = A+ B$ ... eq(1)

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Stille have 2 unkwons, A and B. To use the other initial condition, $\\frac{dy}{dx}=2$ when $x=0$, the general solution must be differentiated.

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\\[\\frac{dy}{dx} = \\var{Rootalpha}Ae^{\\var{Rootalpha}x}+\\var{Rootbeta}Be^{\\var{Rootbeta}x} \\]

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Sub in the conditions

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\\[2 = \\var{Rootalpha}Ae^{\\var{Rootalpha}(0)}+\\var{Rootbeta}Be^{\\var{Rootbeta}(0)} \\]

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\\[2 = \\var{Rootalpha}A+\\var{Rootbeta}B \\]  ....eq(2)

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Let $B=\\var{c}-A$ .... from eq(1)

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Sub into eq(2)

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\\[2 = \\var{Rootalpha}A+\\var{Rootbeta}(\\var{c}-A) \\]

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\\[ A = \\frac{2-(\\var{Rootbeta})(\\var{c})}{\\var{Rootalpha}-\\var{Rootbeta}}= \\var{(2-({Rootbeta})({c}))/({Rootalpha}-{Rootbeta})} \\]

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Sub A into eq(1)

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\\[ \\var{c} = \\var{(2-({Rootbeta})({c}))/({Rootalpha}-{Rootbeta})} + B \\]

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\\[ B= \\var{{c}- (2-({Rootbeta})({c}))/({Rootalpha}-{Rootbeta})} \\]

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Sun back into the general solution

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\\[y = \\var{(2-({Rootbeta})({c}))/({Rootalpha}-{Rootbeta})}e^{\\var{Rootalpha}x}+\\var{{c}- (2-({Rootbeta})({c}))/({Rootalpha}-{Rootbeta})}e^{\\var{Rootbeta}x} \\]...particular solution

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Evaluate the roots for the equation given.

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$\\alpha =$ [[0]]   and    $\\beta =$ [[1]]

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Hint: assign alpha to the answer of $-b+\\sqrt{}$ and $\\beta$ to the $-b - \\sqrt{}$

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What type of roots are these?

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State the general solution to the equation using the roots 

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\\[\\alpha = \\var{Rootalpha}\\] and \\[ \\beta = \\var{Rootbeta}\\]

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y=[[0]]

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Evaluate the constants A and B for the following initial conditions

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$y=\\var{c}$ and $\\frac{dy}{dx}=2$ when $x=0$.

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$ y =$ [[0]] $e^{\\var{Rootalpha}x}+$ [[1]]$e^{\\var{Rootbeta}x}  $

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