Question

Find the differential equation of the family of curves y = e^x (Acos x + Bsin x) where A and B are arbitrary constants and hence solve the equation.

Answer 1

Answer on Question #50669 - Math - Differential calculus | Equations

Question

Find the differential equation of the family of curves $y = e^{x}(A\cos x + B\sin x)$ where $A$ and $B$ are arbitrary constants and hence solve the equation.

Solution

Method 1

From the theory of ordinary differential equations the following fact is known:

if $y = e^{x}(A\cos x + B\sin x)$ is the solution to differential equation, then the characteristic equation has complex roots $\lambda_{1,2} = 1 \pm i$. We obtain characteristic equation of the original differential equation: $(\lambda - 1 - i)(\lambda - 1 + i) = 0 \Leftrightarrow (\lambda - 1)^{2} = -1 \Leftrightarrow \lambda^{2} - 2\lambda + 2 = 0$

Therefore the original differential equation is $y'' - 2y' + 2y = 0$.

If we want to solve it, then we must repeat all previous steps.

Let's construct characteristic equation of the original differential equation: $\lambda^2 - 2\lambda + 2 = 0$. It is easy to see that solutions of the characteristic equation are $\lambda_{1,2} = 1 \pm i$. Since $\lambda_{1,2} = 1 \pm i$ are complex numbers with real part 1 and image part $\pm 1$ then the family of curves $y = e^{x}(A\cos x + B\sin x)$, where $A$ and $B$ are arbitrary real constants, is solution of differential equation $y'' - 2y' + 2y = 0$.

Method 2

To find the differential equation of the family of curves, differentiate $y = e^{x}(A\cos x + B\sin x)$:

$$y' = e^{x}(A\cos x + B\sin x - A\sin x + B\cos x - e^{x}\cos x(A + B) + e^{x}\sin x(B - A)$$ $$y'' = e^{x}(A\cos x + B\sin x - 2A\sin x + 2B\cos x - A\cos x - B\sin x) = 2Be^{x}\cos x - 2Ae^{x}\sin x$$

Suppose that the family of curves $y = e^{x}(A\cos x + B\sin x)$ satisfies equation $y' + \mu y = 0$, then $\frac{A + B}{A} = \frac{B - A}{B}$, which gives infinitely many cases $A^2 + B^2 = 1$, because $A$ and $B$ are arbitrary constants, so we failed to construct the first-order linear differential equation with constant coefficients.

Solve the equation $y'' + \alpha y' + \beta y = 0$ for $\alpha, \beta$:

$$e^{x}\cos x(2B + \alpha(A + B) + \beta A) + e^{x}\sin x(-2A + \alpha(B - A) + \beta B) = 0$$

for arbitrary real $x$, hence obtain the system of equations

$$\left\{ \begin{array}{l} (A + B)\alpha + A\beta = -2B \\ (B - A)\alpha + B\beta = 2A \end{array} \right.$$ $$\Delta = \left| \begin{array}{cc} A + B & A \\ B - A & B \end{array} \right| = AB + B^2 - AB + A^2 = A^2 + B^2, \quad \Delta_1 = \left| \begin{array}{cc} -2B & A \\ 2A & B \end{array} \right| = -2B^2 - 2A^2 = -2(A^2 + B^2),$$ $$\Delta_2 = \left| \begin{array}{cc} A + B & -2B \\ B - A & 2A \end{array} \right| = 2A^2 + 2AB + 2B^2 - 2AB = 2(A^2 + B^2).$$

By Cramer's rule, obtain the unique solution $\alpha = \frac{\Delta_1}{\Delta} = \frac{-2(A^2 + B^2)}{A^2 + B^2} = -2$,

$$\beta = \frac{\Delta_2}{\Delta} = \frac{2(A^2 + B^2)}{A^2 + B^2} = 2.$$

Finally the differential equation will be $y'' - 2y' + 2y = 0$.

**Answer**: $y'' - 2y' + 2y = 0$.

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