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Question

Find a real general solution. Show the details of your work. x²y''+0.7xy'-0.1y=0

Solution

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Let's substitute:

y=xm,y=mxm1,y=m(m1)xm2y = x^m, \quad y' = mx^{m-1}, \quad y'' = m(m-1)x^{m-2}

into the given ODE. This gives:

x2m(m1)xm2+0.7xmxm10.1xm=0x^2m(m-1)x^{m-2} + 0.7 x mx^{m-1} -0.1 x^m = 0

x2m(m1)xmx2+0.7xmxmx10.1xm=0\cancel{x^2}m(m-1)x^m \cdot \cancel{x^{-2}} + 0.7 \cancel{x} mx^{m}\cancel{ x^{-1}} -0.1 x^m = 0

We can see that xmx^m is a common factor, dropping it gives:

m(m1)+0.7m0.1=0    m20.3m0.1=0()m(m-1) + 0.7m -0.1 = 0 \iff m^2 - 0.3m - 0.1 = 0 \quad \textcolor{Fuchsia}{(*)}

So, y=xmy = x^m is a solution of the given ODE if mm is a root of the equation ()\textcolor{Fuchsia}{(*)}

Let's find the roots of the equation ()\textcolor{Fuchsia}{(*)}.

m20.3m0.1=0    m1/2=0.3±0.32+40.12    m1/2=0.3±0.09+0.42    m1/2=0.3±0.72\begin{align*} m^2 - 0.3m -0.1 = 0 &\iff m_{1/2} = \frac{0.3 \pm \sqrt{0.3^2 + 4\cdot 0.1 }}{2} \\ &\iff m_{1/2} = \frac{0.3 \pm \sqrt{0.09 + 0.4}}{2} \\ &\iff m_{1/2} = \frac{0.3 \pm 0.7}{2} \end{align*}

So, it has the distinct real roots:

m1=0.5m2=0.2m _1 = 0.5 \quad \wedge \quad m _2 = -0.2

Real different roots m1m_1 and m2m_2 provide two real solutions:

y1=xm1=x0.5y2=xm2=x0.2y_1 = x^{m_1} = x^{0.5} \quad \wedge \quad y_2 = x^{m_2} = x^{-0.2}

Their quotient is not constant, so the solutions y1y_1 and y2y_2 are linearly independent and constitute a basis of solutions for the given ODE, for all x for which y1,y2Ry_1,y_2 \in \mathbb{R}.

So, the general solution is:

y=c1y1+c2y2=c1x0.5+c2x0.2{\color{#4257b2}{y}}= c_1 y_1 + c_2 y_2 = \boxed{{\color{#4257b2}{c_1x^{0.5} + c_2 x^{-0.2} }}}

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