Question

Solve the differential equation by variation of parameters, subject to the initial conditions
y(0) = 1, y'(0) = 0.

9y'' − y = x(e^x/3)

y(x) = ____

Answer 1

The auxiliary equation is $9m^2+1=0$, which gives $m = \pm\dfrac{1}{3}.$

The complementary function is $C.F = c_{1}e^{-\frac{x}{3}}+c_{2}e^{\frac{x}{3}}$ .

The particular integral is given by,

$P.I = Pf_{1}+Qf_{2}$ , where $f_{1} = e^{-\frac{x}{3}}, f_{2} = e^{\frac{x}{3}}$ and

$P=-\displaystyle\int \dfrac{f_{2}X}{W(f_{1},f_{2})}dx$, $Q=\displaystyle\int \dfrac{f_{1}X}{W(f_{1},f_{2})}dx~ \text{and}~ X = xe^{\frac{x}{3}}$.

Now,

$W(f_{1},f_{2}) = f_{1}f'_{2}-f'_{1}f_{2} \\ ~~~~~~~~~~~~~~~~~~= e^{-\frac{x}{3}} \cdot \frac{e^{\frac{x}{3}}}{3} - \frac{-e^{-\frac{x}{3}}}{3} \cdot e^{\frac{x}{3}} = \frac{2}{3}$

$P=-\displaystyle\int \dfrac{f_{2}X}{W(f_{1},f_{2})}dx =-\displaystyle\int \dfrac{e^{\frac{x}{3}} \cdot xe^{\frac{x}{3}}}{\frac{2}{3}}dx\\ ~~~~=-\dfrac{3}{2}\displaystyle\int xe^{\frac{2x}{3}}dx =-\dfrac{3}{2}\biggl\{\frac{xe^{\frac{2x}{3}}}{\frac{2}{3}} - \dfrac{e^{\frac{2x}{3}}}{\frac{4}{9}}\biggr\}\\ ~~~~=\dfrac{3}{2}\biggl\{\dfrac{9}{4}{e^{\frac{2x}{3}}} - \dfrac{3}{2}{xe^{\frac{2x}{3}}} \biggr\}=\dfrac{9}{8}e^{\frac{2x}{3}}(3-2x)$

$Q=\displaystyle\int \dfrac{f_{1}X}{W(f_{1},f_{2})}dx = \displaystyle\int \dfrac{e^{-\frac{x}{3}} \cdot xe^{\frac{x}{3}}}{\frac{2}{3}}dx = \dfrac{3}{4}x^{2}$

Therefore,

$P.I = Pf_{1}+Qf_{2} \\~~~~~~~= \dfrac{9}{8}e^{\frac{2x}{3}}(3-2x) \cdot e^{-\frac{x}{3}} + \dfrac{3}{4}x^{2} \cdot e^{\frac{x}{3}} \\~~~~~~~~~=\dfrac{3}{8}e^{\frac{x}{3}}(2x^{2}-6x+9)$

Thus,

$y = c_{1}e^{-\frac{x}{3}} + c_{2}e^{\frac{x}{3}} + \dfrac{3}{8}e^{\frac{x}{3}}(2x^{2}-6x+9)$

Given the initial condition, $y(0) = 1, y'(0)=0.$

Using these conditions we get,

$c_{1}+c_{2} = -\frac{19}{8}\\~ -c_{1}+c_{2}=\frac{27}{8}$

Solving, we get $c_{1} = -\frac{23}{8}, c_{2}=\frac{1}{2}$ and hence

$y = -\dfrac{23}{8}e^{-\frac{x}{3}} + \dfrac{1}{2}e^{\frac{x}{3}} + \dfrac{3}{8}e^{\frac{x}{3}}(2x^{2}-6x+9)\\ ~\\~~~=\dfrac{1}{8}e^{\frac{x}{3}}(6x^{2}-18x+31)-\dfrac{23}{8}e^{-\frac{x}{3}}$