The standard form of a linear first order differential equation is \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\). The equation contains the variable \(y\) and its first derivative. \(P(x)\) and \(Q(x)\) are either numeric constants or function of \(x\). It is generally not variable separable.
To solve for&nbsp;\(y\), we first let&nbsp;\(I(x)\) be the integrating factor that will help us solve the equation.
Multiply throughout by \(I(x)\) and we get
\[I(x)\frac{\mathrm{d}y}{\mathrm{d}x} + I(x)P(x)y = I(x)Q(x)\]
Now, if we choose \(I(x)\) such that&nbsp;\(I'(x) = I(x)P(x)\), note that the equation can be re-written as
\[I(x)\frac{\mathrm{d}y}{\mathrm{d}x} + I'(x)y = I(x)Q(x)\]
\[\Rightarrow \frac{\mathrm{d}}{\mathrm{d}x}(I(x) \cdot y) = I(x)Q(x) \text{ by using Product Rule of Differentiation in reverse order}\]
\[\Rightarrow I(x)y = \int I(x)Q(x) \, \mathrm{d}x\]
And hence&nbsp;\(y = \frac{1}{I(x)} \int I(x)Q(x) \, \mathrm{d}x\).
This is of course provided we know what the integrating factor \(I(x)\) is. To find \(I(x)\), consider
\[I'(x) = I(x)P(x) \quad \text{(by construction)}\]
\[\Rightarrow \frac{I'(x)}{I(x)} = P(x)\]
\[\Rightarrow \int \frac{I'(x)}{I(x)} \, \mathrm{d}x = \int P(x) \, \mathrm{d}x\]
\[\Rightarrow \ln|I(x)| = \int P(x) \, \mathrm{d}x\]
\[\Rightarrow I(x) = e^{\int P(x) \, \mathrm{d}x} \quad \text{or} \quad -e^{\int P(x) \, \mathrm{d}x}\]
However we can just take \(I(x) = e^{\int P(x) \, \mathrm{d}x}\) without loss of generality (because of symmetry).
Summary
To solve a first order linear differential equation of the form \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\),
<ol class="num-list"><li><div class="list-item-content">Find the integrating factor \(I(x)\) using the formula \(I(x) = e^{\int P(x) \, \mathrm{d}x}\)</div></li><li><div class="list-item-content">Multiply throughout by the integrating factor and observe that the LHS can be re-written as the derivative of the product \(I(x)y\)</div></li><li><div class="list-item-content">Integrate both sides directly.</div></li></ol>

The standard form of a linear first order differential equation is \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\). The equation contains the variable \(y\) and its first derivative. \(P(x)\) and \(Q(x)\) are either numeric constants or function of \(x\). It is generally not variable separable.

To solve for \(y\), we first let \(I(x)\) be the **integrating factor** that will help us solve the equation.

Multiply throughout by \(I(x)\) and we get

\[I(x)\frac{\mathrm{d}y}{\mathrm{d}x} + I(x)P(x)y = I(x)Q(x)\]

Now, if we choose \(I(x)\) such that \(I'(x) = I(x)P(x)\), note that the equation can be re-written as

\[I(x)\frac{\mathrm{d}y}{\mathrm{d}x} + I'(x)y = I(x)Q(x)\]

\[\Rightarrow \frac{\mathrm{d}}{\mathrm{d}x}(I(x) \cdot y) = I(x)Q(x) \text{ by using Product Rule of Differentiation in reverse order}\]

\[\Rightarrow I(x)y = \int I(x)Q(x) \, \mathrm{d}x\]

And hence \(y = \frac{1}{I(x)} \int I(x)Q(x) \, \mathrm{d}x\).

This is of course provided we know what the integrating factor \(I(x)\) is. To find \(I(x)\), consider

\[I'(x) = I(x)P(x) \quad \text{(by construction)}\]

\[\Rightarrow \frac{I'(x)}{I(x)} = P(x)\]

\[\Rightarrow \int \frac{I'(x)}{I(x)} \, \mathrm{d}x = \int P(x) \, \mathrm{d}x\]

\[\Rightarrow \ln|I(x)| = \int P(x) \, \mathrm{d}x\]

\[\Rightarrow I(x) = e^{\int P(x) \, \mathrm{d}x} \quad \text{or} \quad -e^{\int P(x) \, \mathrm{d}x}\]

However we can just take \(I(x) = e^{\int P(x) \, \mathrm{d}x}\) without loss of generality (because of symmetry).

To solve a first order linear differential equation of the form \(\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)\),

Find the integrating factor \(I(x)\) using the formula \(I(x) = e^{\int P(x) \, \mathrm{d}x}\)

Multiply throughout by the integrating factor and observe that the LHS can be re-written as the derivative of the product \(I(x)y\)

Differential Equations

International Baccalaureate Higher Level Mathematics

First order linear equations

First order linear equations