This problem aims to familiarize us with the **method** of **variation** of **parameters.** The concepts required for this problem are related to **ordinary differential equations** which include **general, particular, fundamental solutions** and** the Wronskian.**

We will start by looking at **variation of parameters** which deals with the **equation** of the form $\dfrac{d^2y}{dx^2} + p\dfrac{dy}{dx} + qy = f(x)$.

The **complete solution** can be found using a **combination** of the following methods:

- – The
**general solution**of $\dfrac{d^2y}{dx^2} + p\dfrac{dy}{dx} + qy = 0$ (**homogeneous equation**). - –
**Particular solutions**of $\dfrac{d^2y}{dx^2} + p\dfrac{dy}{dx} + qy = f(x)$ (**non-homogeneous equation**).

The **complete solution** can thus be found by adding all the solutions. This approach depends on **integration.**

Whereas the **Wronksian** is found when $y_1$ and $y_2$ are the **two solutions** of the **homogeneous** equation:

$W(y_1,y_2) = y_1\space y_2`\space -\space y_2\space y_1`$, where $y_1$ and $y_2$ are **independent.**

## Expert Answer

The given **equation** is:

\[ y“ + y = sinx \]

The **characteristics equation** for this equation is $r^2 + 1 = 0$, which has **roots** $r = \pm i$.

The **complementary solution** of the equation can be found by taking the **integral** of the main equation:

\[\int y“ d(x) +\int y dx =\int sinx dx\]

\[ y_c = C_1cosx + C_2sinx\]

This **complementary solution** is split into two **independent** solutions as:

\[ y_1 = cosx \space \space y_2 = sinx\]

Then we can find the **Wronksian** as:

\[ W(y_1,y_2) = \begin{bmatrix} cosx & sinx \\ -sinx & cosx \end{bmatrix} \]

\[ W(y_1,y_2) = cos^2x + sin^2x \]

Using the **trigonometric** identity:

\[ W(y_1,y_2) = 1 \]

Now, **solving** for $W_1$:

\[ W_1 = \begin{bmatrix} 0 & sinx \\ sinx & cosx \end{bmatrix} \]

\[ W_1 = -sin^2x\]

\[ W_1 = \dfrac{1-cos2x}{2}\]

\[ W_1 =\dfrac{-1}{2} + \dfrac{1}{2}cos2x\]

Now, **solving** for $W_2$:

\[W_2 = \begin{bmatrix} cosx & 0 \\ -sinx & sinx \end{bmatrix} \]

\[W_2 = sinx + cosx \]

\[W_2 = \dfrac{1}{2}(2sinxcosx) \]

\[W_2 = \dfrac{1}{2}(sin2x) \]

The **particular solution** is given by the equation $y_p = u_1y_1 + u_2y_2$ found by the **integration:**

\[u_1 = \int \dfrac{W_1}{W} dx\]

\[= \int \dfrac{\dfrac{-1}{2} + \dfrac{1}{2}cos2x}{1} dx\]

\[= \dfrac{-1}{2}\int dx + \dfrac{1}{2}\int cos2x dx\]

\[u_1= -\dfrac{1}{2}x + \dfrac{1}{4}sin2x\]

Now **finding** $u_2$:

\[u_2 = \int \dfrac{W_2}{W} dx\]

\[= \int \dfrac{\dfrac{1}{2} sin2x}{1} dx\]

\[= \dfrac{1}{2}\int sin2x dx\]

\[u_2= -\dfrac{1}{4}cos2x\]

**Plugging** the values:

\[y_p=\dfrac{-1}{2}xcosx + \dfrac{1}{4}sin2xcosx – \dfrac{1}{4}cos2xsinx\]

Now the **general solution** is the **combination** of all the solutions:

\[y=y_c + y_p\]

\[y=C_1cosx + C_2sinx – \dfrac{1}{2}xcosx + \dfrac{1}{4}sin2xcosx – \dfrac{1}{4}cos2xsinx\]

## Numerical Result

The **general solution** comes out to be:

\[y=C_1cosx + C_2sinx – \dfrac{1}{2}xcosx + \dfrac{1}{4}sin2xcosx – \dfrac{1}{4}cos2xsinx\]

## Example

Without **solving,** specify the **Wronskian** value of $2$ **solutions** for:

$t^4y“ – 2t^3y` – t^8y = 0$

The first thing to do here is to **divide** this **differential equation** by the **coefficient** of the highest derivative as it will yield the solution. This will give us:

\[ y“ – \dfrac{2}{t}y` – t^4y = 0\]

Now using the **equation:**

\[W(y_1,y_2) \space (t) = ce^{-\int p(t) dt}\]

\[= ce^{-\int – \dfrac{2}{t} dt}\]

\[= ce^{2\ln t}\]

\[=ce^{\ln t^2}\]

\[ W = ct^2\]