Question

solve: d^2/dx^2 - 2 tan x dy/dx + 5y = e^x sec x

Answer 1

Answer on Question #70345 - Math - Calculus

Solve: $\frac{d^2y}{dx^2} - 2\tan x\frac{dy}{dx} + 5y = e^x\sec x$

Solution

We are given

$$\frac{d^2y}{dx^2} - 2\tan x\frac{dy}{dx} + 5y = e^x\sec x$$

Multiply both sides of the equation by $\cos x$:

$$\cos x \frac{d^2y}{dx^2} - 2\sin x \frac{dy}{dx} + 5y\cos x = e^x$$

We will be looking for a solution in the form

$$u(x) = y(x)\cos x$$

Find $\frac{du}{dx}$ and $\frac{d^2u}{dx^2}$

$$\frac{du}{dx} = \frac{dy}{dx}\cos x - y\sin x$$ $$\frac{d^2u}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\cos x - y\sin x\right) = \frac{d^2y}{dx^2}\cos x - \frac{dy}{dx}\sin x - \frac{dy}{dx}\sin x - y\cos x = \frac{d^2y}{dx^2}\cos x - 2\frac{dy}{dx}\sin x - y\cos x = \frac{d^2y}{dx^2}\cos x - 2\frac{dy}{dx}\sin x - u$$

Thus, from the last equality we get

$$\frac{d^2y}{dx^2}\cos x - 2\frac{dy}{dx}\sin x = \frac{d^2u}{dx^2} + u$$

Substitution of the (3) into the (1) yields

$$\frac{d^2u}{dx^2} + 6u = e^x$$

Solve this equation. The general solution is the sum of the complementary solution and particular solution. Find the complementary solution by solving

$$\frac{d^2u}{dx^2} + 6u = 0$$

Substitute $u = e^{\lambda x}$

$$\frac{d^2}{dx^2}(e^{\lambda x}) + 6e^{\lambda x} = 0$$ $$\lambda^2 e^{\lambda x} + 6e^{\lambda x} = 0 \Rightarrow (\lambda^2 + 6)e^{\lambda x} = 0 \Rightarrow \lambda^2 + 6 = 0 \Rightarrow \lambda^2 = -6 \Rightarrow \lambda_1 = i\sqrt{6}, \lambda_2 = -i\sqrt{6}$$

The obtained roots give $u_{1}(x) = c_{1}e^{i\sqrt{6} x}$, $u_{2}(x) = c_{2}e^{-i\sqrt{6} x}$ as solutions, where $c_{1}$ and $c_{2}$ are arbitrary constants. The general solution is the sum of $u_{1}$ and $u_{2}$:

$$u(x) = u_{1}(x) + u_{2}(x) = c_{1}e^{i\sqrt{6}x} + c_{2}e^{-i\sqrt{6}x}$$

Apply Euler's identity $e^{\alpha + i\beta} = e^{a}\cos\beta + i e^{a}\sin\beta$

$$\begin{aligned} u(x) = c_{1}e^{i\sqrt{6}x} + c_{2}e^{-i\sqrt{6}x} &= c_{1}\left(\cos(\sqrt{6}x) + i\sin(\sqrt{6}x)\right) + c_{2}\left(\cos(\sqrt{6}x) - i\sin(\sqrt{6}x)\right) = \\ &= (c_{1} + c_{2})\cos(\sqrt{6}x) + i(c_{1} - c_{2})\sin(\sqrt{6}x) \end{aligned}$$

Redefine $(c_{1} + c_{2})$ as $c_{1}$ and $i(c_{1} - c_{2})$ as $c_{2}$, since these are arbitrary constants:

$$= c_{1}\cos(\sqrt{6}x) + c_{2}\sin(\sqrt{6}x)$$

Determine the particular solution to $\frac{d^2u}{dx^2} + 6u = e^x$ by the method of undetermined coefficients:

The particular solution to $\frac{d^2u}{dx^2} + 6u = e^x$ is of the form

$$u_{p}(x) = a_{1}e^{x}$$

Substitute the particular solution $u_{p}(x)$ into the differential equation:

$$\frac{d^2}{dx^2}(a_{1}e^{x}) + 6a_{1}e^{x} = e^{x}$$ $$a_{1}e^{x} + 6a_{1}e^{x} = e^{x}$$ $$a_{1} + 6a_{1} = 1 \Rightarrow 7a_{1} = 1 \Rightarrow a_{1} = \frac{1}{7}$$

Substitute $a_{1}$ into $u_{p}(x) = a_{1}e^{x}$

$$u_{p}(x) = \frac{1}{7}e^{x}$$

The general solution of (4) is

$$u(x) = u_{c}(x) + u_{p}(x) = \frac{1}{7}e^{x} + c_{1}\cos(\sqrt{6}x) + c_{2}\sin(\sqrt{6}x)$$

We find solution of the original equation from the equality (2)

$$u(x) = y(x)\cos x \Rightarrow y(x) = \frac{1}{\cos x} u(x) = \frac{1}{\cos x}\left(\frac{1}{7}e^{x} + c_{1}\cos(\sqrt{6}x) + c_{2}\sin(\sqrt{6}x)\right)$$

Answer: $y(x) = \frac{1}{\cos x}\left(\frac{1}{7}e^{x} + c_{1}\cos(\sqrt{6}x) + c_{2}\sin(\sqrt{6}x)\right)$

Answer provided by https://www.AssignmentExpert.com