Chapter 4 - In Review

1. $y = 0$

2. Since $y_c = c_1e^x + c_2e^{-x}$, a particular solution for $y’’ — y = 1 + e^x}$ is $y_p = A+ Bxe^x$.

3. It is not true unless the differential equation is homogeneous. For example, $y_1= x$ is a solution $y" + y = x$, but $y_2 = 5x$ is not.

4. True

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15. From $m^2-2m-2=0$ we obtain $m=1 \pm \sqrt{3}$ so that $y=c_1e^{(1+\sqrt{3})x}+ c_2e^{(1-\sqrt{3})x}$.

16. From$2m^2+2m+3=0$ we obtain $m=-\frac{1}{2} \pm \frac{\sqrt{5}}{2}i$ so that $y=e^{-\frac{x}{2}}\left(c_1\, cos \frac{\sqrt{5}}{2}+ c_2 \,sin\frac{\sqrt{5}}{2}x \right)$.

17. From $m^3 + 10m^2 + 25m = 0$ we obtain $m = 0$, $m= — 5$, and $m = —5$ so that $y = c_1 + c_2e^{-5x} + c_3xe^{-x}$.

18. From $2m^3 + 9m^2 + 12m + 5 = 0$ we obtain $m = —1$, $m = —1$, and $m = —\frac{5}{2}$ so that $y = c_1e^{-\frac{5x}{2}} + c_2e^{-x} + c_3xe^{-x}$.

19. From $3m^3 + 10m^2 + 15m + 4 = 0$ we obtain $m = —\frac{1}{3}$ and $m = —\frac{3}{2} \pm \left(\frac{\sqrt{7}}{2}\right)i$ so that $y = c_1e^{-\frac{x}{3}} + e^{-2x}\left( c_2\,cos\, \frac{\sqrt{7}}{2}x + c_3 \,sin \frac{\sqrt{7}}{2}x\right)$.

20. From $2m^4 + 3m^3 + 2m^2 + 6m - 4 = 0$ we obtain $m= \frac{1}{2}$, $m = -2$, and $m = \pm \sqrt{2}\,i$ so that $y = c_1e^{\frac{x}{2}} + c_2e^{-2x} + c_3 \,cos \sqrt{2}\, x + c_4\, sin \,\sqrt{2}\,x$.

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27. The auxiliary equation is $6m^2 - m - 1 = 0$ so that $y = c_1x^{\frac{1}{2}} + c_2x^{-frac{1}{3}}$.

28. The auxiliary equation is $2m^3 + 13m^2 + 24m + 9 = (m + 3)^2\left(m + \frac{1}{2}\right) = 0$ so that $y = c_1x^{-3} + c_2x^{-3}\, In \,x + c_3x^{-\frac{1}{2}}$.

29. The auxiliary equation is $m^2 - 5m + 6 = (m - 2) (to - 3) = 0$ and a particular solution is $y_p= x^4 - x^2\, ln\, x$ so that $y =c_1x^2+c_2x^3 + x^4 - x^2\, ln\, x$.

30. The auxiliary equation is $m^2 - 2m + 1= (m - 1 )^2 = 0$ and a particular solution is $y_p = \frac{1}{4}x^3$ so that $y = c_1x + c_2x \,ln \,x + \frac{1}{4}x^3$.

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36. The auxiliary equation is $m^2+2m+1 = (m+1)^2 = 0$, so that $y = c_1e^{-x}+c_2xe^{-x}$. Setting $y(-1) = 0$ and $y'(0) = 0$ we get $c_1e - c_2e = 0$ and $-c_1 + c_2 = 0$. Thus $c_1 = c_2$ and $y =c_1(e^{-x} + xe^{-x})$ is a solution of the boundary-value problem for any real number $c_1$.

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