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Question

(a) Verify that the given functions are linearly independent and form a basis of solutions of the given ODE. (b) Solve the IVP. Graph or sketch the solution.

y”+0.6y’+0.09y=0, y(0)=2.2, y’(0)=0.14, e0.3xxe0.3x\text{y''+0.6y'+0.09y=0, y(0)=2.2, y'(0)=0.14, }e^{-0.3x}\text{, }xe^{-0.3x}

Solution

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4.4 (8 ratings)
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We have

y+0.6y+0.09y=0,  y(0)=2.2,  y(0)=0.14,  e0.3x,  xe0.3xy''+0.6y'+0.09y=0, \; y\left( 0\right)=2.2, \; y'\left( 0\right)=0.14, \; e^{-0.3x}, \; xe^{-0.3x}

where,

y1y2=e0.3xxe0.3x=1xconst\dfrac{y_{1}}{y_{2}}=\dfrac{\bcancel{e^{-0.3x}}}{x\bcancel{e^{-0.3x}}}=\dfrac{1}{x} \neq \text{const}

and we see that  e0.3x  and  xe0.3x  are linearly independent on any interval.\text{and we see that}\; e^{-0.3x} \; \text{and} \; xe^{-0.3x}\; \text{are linearly independent on any interval.}

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