Method for Solving Linear Equations
- Write the equation in standard form
$$
\frac{dy}{dx} + P(x) y = Q(x).
$$ - Calculate the integrating factor \(\mu(x)\) by the formula
$$
\mu(x) = e^{\int P(x) \; dx}.
$$ - Multiply the equation in standard form by \(\mu(x)\) and, recalling that the left-hand side is just \( \frac{d}{dx}[\mu(x) y ], \) obtain
$$
\begin{array}{c c c}
\underbrace{\mu(x) \frac{dy}{dx} + P(x) \mu(x) y} & = & \mu(x) Q(x), \\
\frac{d}{dx}\left[ \mu(x) y \right] & = & \mu(x) Q(x).
\end{array}
$$ - Integrate the last equation and solve for \(y\) by dividing by \(\mu(x)\) to obtain
$$
y(x) = \frac{1}{\mu(x)} \left[ \int \mu(x) Q(x) \; dx + C \right].
$$
Existence and Uniqueness of Solution
Suppose \(P(x)\) and \(Q(x)\) are continuous on an interval \((a,b)\) that contains the point \(x_0\). Then for any choice of initial values \(y_0,\) there exists a unique solution \(y(x)\) on \((a,b)\) to the initial value problem
$$
\frac{dy}{dx} + P(x) y = Q(x), \quad y(x_0) = y_0.
$$
In fact, the solution is given by
$$
y(x) = \frac{1}{\mu(x)} \left[ \int \mu(x) Q(x) \; dx + C \right],
$$
for some constant \(C\).
Suppose \(P(x)\) and \(Q(x)\) are continuous on an interval \((a,b)\) that contains the point \(x_0\). Then for any choice of initial values \(y_0,\) there exists a unique solution \(y(x)\) on \((a,b)\) to the initial value problem
$$
\frac{dy}{dx} + P(x) y = Q(x), \quad y(x_0) = y_0.
$$
In fact, the solution is given by
$$
y(x) = \frac{1}{\mu(x)} \left[ \int \mu(x) Q(x) \; dx + C \right],
$$
for some constant \(C\).