Question

3. a) Write a suitable form of particular solution for solving the equation
y'' + 3y' + 2y = e^x (x^2 + 1) sin 2x + 3e^-x cos x + 4e^x
by the method of undetermined coefficients.

b) Verify that e^x and xe^x are solutions of the homogeneous equation corresponding to the
equation
y''− 2y'+ y = e^x / (1 + x^2), −&<x<&
and find the general solution using the method of variation of parameters.

c) If y1 = 2x+2 and y2 = -x^2 / 2 are the solutions of
xy' + y' - y'^2 / 2
then are the constant multiples c1 y1 and c2 y2 , where c1 and c2 are arbitrary, also solutions
of the given differential equation? Is the sum y1 + y2 a solution? Justify your answer.

Answer 1

Answer on Question #50894 – Math – Differential Calculus | Equations

a) Write a suitable form of particular solution for solving the equation

$$y'' + 3y' + 2y = e^x(x^2 + 1) \sin 2x + 3e^{-x} \cos x + 4e^x$$

by the method of undetermined coefficients.

b) Verify that $e^x$ and $xe^x$ are solutions of the homogeneous equation corresponding to the equation

$$y'' - 2y' + y = \frac{e^x}{1 + x^2}, \quad -\infty < x < \infty$$

and find the general solution using the method of variation of parameters.

c) If $y_1 = 2x + 2$ and $y_2 = -\frac{x^2}{2}$ are the solutions of

$$xy' + y' - \frac{y'^2}{2}$$

then are the constant multiples $c_1y_1$ and $c_2y_2$, where $c_1$ and $c_2$ are arbitrary, also solutions of the given differential equation? Is the sum $y_1 + y_2$ a solution?

Solution

a) We first consider the solutions to the corresponding homogeneous equation. Using the characteristic equation $m^2 + 3m + 2 = 0 \rightarrow m_1 = -2, m_2 = -1$, we have solutions $y_1 = e^{-2x}$ and $y_2 = e^{-x}$. Thus we have the form

$$Y = (Ax^2 + Bx + C)e^x \sin 2x + (Dx^2 + Ex + F)e^x \cos 2x + Ge^{-x} \cos x + He^{-x} \sin x + Ie^x.$$

b) The homogeneous equation corresponding to the equation

$$y'' - 2y' + y = \frac{e^x}{1 + x^2}, \quad -\infty < x < \infty$$

is

$$y'' - 2y' + y = 0, \quad -\infty < x < \infty$$ $$\frac{d^2}{dx^2}(e^x) = e^x, \quad \frac{d}{dx}(e^x) = e^x.$$

So,

$$y'' - 2y' + y = e^x - 2e^x + e^x \equiv 0.$$

And we see that $e^x$ is a solution.

$$\frac{d^2}{dx^2}(xe^x) = (x + 2)e^x, \quad \frac{d}{dx}(xe^x) = (x + 1)e^x.$$

So,

$$y'' - 2y' + y = (x + 2)e^x - 2(x + 1)e^x + xe^x \equiv 0.$$

And we see that $xe^{x}$ is a solution.

$$y = c_{1} e^{x} + c_{2} x e^{x}.$$

The Wronskian is given by

$$W = \begin{bmatrix} e^{x} & x e^{x} \\ e^{x} & (x + 1) e^{x} \end{bmatrix} = (x + 1) e^{2x} - x e^{2x} = e^{2x} \neq 0.$$

We seek the solution in the form $y = y_{1} v_{1} + y_{2} v_{2}$

$$v_{1} = \int \left(- \frac {y_{2} R}{W}\right) d x = - \int \frac {x e^{x} \cdot \frac {e^{x}}{1 + x^{2}}}{e^{2x}} d x = \ln \frac {1}{\sqrt {1 + x^{2}}} + c_{1}.$$ $$v_{2} = \int \left(\frac {y_{1} R}{W}\right) d x = \int \frac {e^{x} \cdot \frac {e^{x}}{1 + x^{2}}}{e^{2x}} d x = \tan^{-1} x + c_{2}.$$

So,

$$y = e^{x} \left[ \ln \frac {1}{\sqrt {1 + x^{2}}} + c_{1} \right] + x e^{x} [\tan^{-1} x + c_{2}].$$

c) The constant multiples $c_{1}y_{1}$ and $c_{2}y_{2}$ or the sum $y_{1} + y_{2}$ are solutions if and only if an equation is a linear. But

$$x y' + y' - \frac {y'^{2}}{2}$$

contains nonlinear element (quadratic $\frac{y'^{2}}{2}$). Thus $c_{1}y_{1}$ and $c_{2}y_{2}$ or the sum $y_{1} + y_{2}$ are not solutions.

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