Solve the following differential equation.

{eq}\displaystyle (\sin x) y' + (\cos x) y = \cos 2x {/eq}

Question:

Solve the following differential equation.

{eq}\displaystyle (\sin x) y' + (\cos x) y = \cos 2x {/eq}

Product Rule in Differentiation

Consider a differentiable function {eq}f(x) {/eq} that can be expressed as a product of two differentiable functions {eq}g(x) {/eq} and {eq}h(x) {/eq}:

$$\begin{align} f(x) = g(x) h(x) \\ \end{align} $$

The derivative of {eq}f(x) {/eq} can be written using the Product Rule:

$$\begin{align} f'(x) = g(x) h'(x) + g'(x)h(x) \\ \end{align} $$

Conversely, an expression of the form {eq}g(x) h'(x) + g'(x)h(x) {/eq} can be written in a condensed form as {eq}(g(x) \ h(x))' {/eq}. This result is sometimes useful while solving differential equations.

Answer and Explanation: 1

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Given differential equation:

$$\begin{align} &\sin{x} y' + \cos{x} y = \cos{2x} \\ \Rightarrow & \dfrac{dy}{dx} \sin{x} + \cos{x} y = \cos 2x...

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First-Order Linear Differential Equations

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Chapter 16 / Lesson 3
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Learn to define what a linear differential equation and a first-order linear equation are. Learn how to solve the linear differential equation. See examples.


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