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The initial value problem presented is:

\[
\cos(x)y' + \sin(x)y = 2\cos^3(x)\sin(x) - 1
\]
with the condition
\[
y\left(\frac{\pi}{4}\right) = 3\sqrt{2}.
\]

We will solve this first-order linear differential equation using an integrating factor approach. Let's proceed step by step.

### Step 1: Rewrite the equation in standard linear form
The equation is already in the form:

\[
y' + P(x)y = Q(x)
\]
where \(P(x) = \frac{\sin(x)}{\cos(x)} = \tan(x)\) and \(Q(x) = \frac{2\cos^3(x)\sin(x) - 1}{\cos(x)}\).

Thus, the equation becomes:

\[
y' + \tan(x) y = 2\cos^2(x)\sin(x) - \sec(x).
\]

### Step 2: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[
\mu(x) = e^{\int P(x) dx} = e^{\int \tan(x) dx}.
\]

The integral of \(\tan(x)\) is \(\ln|\sec(x)|\), so the integrating factor becomes:

\[
\mu(x) = e^{\ln|\sec(x)|} = \sec(x).
\]

### Step 3: Multiply the entire equation by the integrating factor
Multiplying both sides of the equation by \(\sec(x)\), we get:

\[
\sec(x)y' + \sec(x)\tan(x)y = 2\cos^2(x)\sin(x) - \sec(x).
\]

The left side simplifies to:

\[
\frac{d}{dx} \left( y \sec(x) \right).
\]

Thus, the equation becomes:

\[
\frac{d}{dx} \left( y \sec(x) \right) = 2\cos^2(x)\sin(x)

Solve the initial value problem cos(x)y' + sin(x)y = 2cos^3(x)sin(x) - 1 with the condition y(π/4) = 3√2.

Learn how to solve the first-order linear differential equation cos(x)y' + sin(x)y = 2cos^3(x)sin(x) - 1 using the integrating factor method. This step-by-step solution involves key mathematical concepts like trigonometric functions and integrating factors. The initial condition y(π/4) = 3√2 is also applied.

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