If \( r_{1} \) and \( r_{2} \) are the values of \( r \) that you found in part (a), show that every member of the family of functions \( y=a e^{r_{1} x}+b e^{r_{2} x} \) is also a solution.
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In part (a) substituting y = e^{r x} gave the characteristic equation r^2 + p r + q = 0, whose roots are r1 and r2. Show more…
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(a) For what values of $r$ does the function $y=e^{r x}$ satisfy the differential equation $2 y^{\prime \prime}+y^{\prime}-y-0 ?$ (b) If $r_{1}$ and $r_{2}$ are the values of $r$ that you found in part (a), show that every member of the family of functions $y=a e^{r_{1} x}+b e^{r x}$ is also a solution.
(a) For what values of $r$ does the function $y=e^{r x}$ satisfy the differential equation $2 y^{\prime \prime}+y^{\prime}-y=0 ?$ (b) If $r_{1}$ and $r_{2}$ are the values of $r$ that you found in part (a) show that every member of the family of functions $y=a e^{r_{1} x}+b e^{r_{2} x}$ is also a solution.
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