Example 3. Find the general solutions of the given linear homogeneous differential equations with constant coefficients. In those cases where initial conditions are given, find the corresponding particular solutions:

1) $2 y^{\prime \prime}-3 y^{\prime}+y=0$
2) $9 y^{\prime \prime}+12 y^{\prime}+4 y=0 ; x=1, y=2, y^{\prime}=1$;
3) $y^{\prime \prime}+4 y^{\prime}+5 y=0$
4) $4 y^{\prime \prime}+25 y=0, x=\pi, y=2, y^{\prime}=1$.
Solution. All differential equations are solved according to a unified scheme, so minor details are sometimes omitted. We form the characteristic equation, find its roots, and construct the general solution of the homogeneous linear differential equation (LODE) according to the algorithm in point $4^{\circ}$.

1) We have $2 k^{2}-3 k+1=0, \quad k_{1}=1, \quad k_{2}=0.5$.

According to 1) in point $4^{\circ}$, we get

Answer. $y_{00}=C_{1} e^{x}+C_{2} e^{0.5 x}$.
2) $9 k^{2}+12 k+4=0, (3 k+2)^{2}=0, k=-\frac{2}{3}$. According to 2) in point $4^{\circ}$, we get the general solution $y_{00}=\left(C_{1}+C_{2} x\right) e^{-(2 / 3) x}$. To find a particular solution, we need to differentiate this equality: $y_{\mathrm{oo}}^{\prime}=\left[C_{2}-\frac{2}{3}\left(C_{1}+C_{2} x\right)\right] e^{-(2 / 3) x}$.

Substituting the initial conditions $x=1$, $y=2, y^{\prime}=1$ into the last equalities, we get the system $\left\{\begin{array}{c}C_{1}+C_{2}=2 e^{2 / 3}, \\ -\frac{2}{3} C_{1}+\frac{1}{3} C_{2}=e^{2 / 3} .\end{array}\right.$ From this, $\left\{\begin{array}{l}C_{2}=\frac{7}{3} e^{2 / 3}, \\ C_{1}=-\frac{1}{3} e^{2 / 3} .\end{array}\right.$ The desired particular solution is

$$
y_{4}=\frac{1}{3}(7 x-1) e^{(2 / 3)(1-x)}
$$

Answer. $y_{\text {oo }}=\left(C_{1}+C_{2} x\right) e^{-(2 / 3) x} ; y_{4}=\frac{1}{3}(7 x-1) e^{(2 / 3)(1-x)}$.
3) $k^{2}+4 k+5=0, \quad k_{1,2}=-2 \pm i$.

According to 3) in point $4^{\circ}$, we get

Answer. $y_{\text {oo }}=\left(C_{1} \cos x+C_{2} \sin x\right) e^{-2 x}$.
4) $4 k^{2}+25=0, \quad k_{1,2}= \pm 2.5 i$. The general solution of the homogeneous equation is $y_{о 0}=C_{1} \cos 2.5 x+C_{2} \sin 2.5 x$. We differentiate this equality:

$$
y_{\mathrm{oo}}^{\prime}=-2.5 C_{1} \sin 2.5 x+2.5 C_{2} \cos 2.5 x
$$

Substituting the initial conditions $x=\pi$, $y=2, y^{\prime}=1$ into the expressions for $y$ and $y^{\prime}$, we get a system of two linear equations with two unknowns:

$$
\left\{\begin{array}{l}
C_{1} \cdot 0+C_{2}=2 \\
-2.5 C_{1}+0 \cdot C_{2}=1
\end{array}\right.
$$

The solvability of the system is ensured by Theorem 2. We find $C_{1}=-0.4 ; C_{2}=2$.

Answer. $y_{\text {оо }}=C_{1} \cos 2.5 x+C_{2} \sin 2.5 x ; \quad y_{\text {ч }}=-0.4 \cos 2.5 x+$ $+2 \sin 2.5 x$

## Exercises

Find the general solutions of linear homogeneous differential equations of the second order with constant coefficients. Verify that the obtained general solution indeed satisfies the given equation. Where initial conditions are provided, find the particular solutions.

1. $2 y^{\prime \prime}+3 y^{\prime}=0.2 .2 y^{\prime \prime}+3 y^{\prime}-5 y=0 ; x=0, y=\frac{7}{2} ; y^{\prime}=0$.
2. $2 y^{\prime \prime}+3 y^{\prime}+5 y=0$. 4. $y^{\prime \prime}+4 y^{\prime}=0 ; x=0, y=0, y^{\prime}=5$.
3. $y^{\prime \prime}+4 y^{\prime}+4 y=0$. 6. $y^{\prime \prime}+4 y^{\prime}+13 y=0$.
4. $4 y^{\prime \prime}+y^{\prime}=0$. 8. $4 y^{\prime \prime}+4 y^{\prime}+y=0 ; x=1, y=e^{-1 / 2}, y^{\prime}=\frac{3}{2} e^{-1 / 2}$.
5. $2 y^{\prime \prime}+2 y^{\prime}+3 y=0$. 10. $4 y^{\prime \prime}+9 y=0$.

## Answers

1. $y_{00}=C_{1}+C_{2} e^{-(3 / 2) x}$. 2. $y_{00}=C_{1} e^{x}+C_{2} e^{-(5 / 2) x} ; y_{4}=$ $=\frac{5}{2} e^{x}+e^{-(5 / 2) x}$. 3. $y_{00}=e^{-(3 / 4) x}\left(C_{1} \cos \frac{\sqrt{31}}{4} x+C_{2} \sin \frac{\sqrt{31}}{4} x\right)$. 4. $y_{00}=C_{1}+C_{2} e^{-4 x} ; y_{4}=1-e^{-4 x}$. 5. $y_{00}=\left(C_{1}+C_{2} x\right) e^{-2 x}$
2. $y_{\mathrm{o0}}=e^{-2 x}\left(C_{1} \cos 3 x+C_{2} \sin 3 x\right)$. 7. $y_{\mathrm{oo}}=C_{1}+C_{2} e^{-(1 / 4) x}$. 8. $y_{00}=\left(C_{1}+