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3 y^prime prime-y^prime-2 y=0

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3 y^prime prime-y^prime-2 y=0

3 y^prime prime-y^prime-2 y=0

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мастер · Репетитор 5 лет

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To solve the given differential equation \(3y'' - y' - 2y = 0\), we can use the characteristic equation method.<br /><br />Step 1: Write the characteristic equation.<br />The characteristic equation is obtained by substituting \(y = e^{rx}\) into the differential equation. This gives us:<br />\[3r^2 - r - 2 = 0\]<br /><br />Step 2: Solve the characteristic equation.<br />We can solve the quadratic equation \(3r^2 - r - 2 = 0\) using the quadratic formula:<br />\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]<br />where \(a = 3\), \(b = -1\), and \(c = -2\).<br /><br />Substituting these values, we get:<br />\[r = \frac{1 \pm \sqrt{1 + 24}}{6}\]<br />\[r = \frac{1 \pm \sqrt{25}}{6}\]<br />\[r = \frac{1 \pm 5}{6}\]<br /><br />So, the roots are:<br />\[r_1 = \frac{6}{6} = 1\]<br />\[r_2 = \frac{-4}{6} = -\frac{2}{3}\]<br /><br />Step 3: Write the general solution.<br />The general solution to the differential equation is given by:<br />\[y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x}\]<br />where \(C_1\) and \(C_2\) are arbitrary constants.<br /><br />Substituting the values of \(r_1\) and \(r_2\), we get:<br />\[y(x) = C_1 e^x + C_2 e^{-\frac{2}{3}x}\]<br /><br />Therefore, the general solution to the differential equation \(3y'' - y' - 2y = 0\) is:<br />\[y(x) = C_1 e^x + C_2 e^{-\frac{2}{3}x}\]
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