For a linear first-order differential equation y +P(x)y=Q(x), work on an interval where the coefficient functions are continuous. First divide by the coefficient of y, if necessary, to obtain this standard form. Method Choose the integrating factor I(x)=e (x),dx. Any one antiderivative of P may be used. Iy +IPy=IQ ddx(Iy)=IQ Iy=,dx+C y= 1I(,dx+C). Use an initial condition to determine C. Check the resulting solution in the original equation and keep the interval on which that equation is defined. Why the method works The chain rule gives I =PI. Therefore the product rule makes the left side the derivative of Iy: ddx(Iy)=Iy +I y=I(y +Py). To obtain the factor, solve I I=P. Integration gives ln|I|=,dx+c, so every nonzero constant multiple of e,dx is an integrating factor. Multiplying the factor by a nonzero constant multiplies the whole transformed equation by that constant. It does not change the solution family; choose the positive exponential factor for convenience.

For a linear first-order differential equation \(y^{\prime}+\mathrm{P}(x)y=\mathrm{Q}(x)\), work on an interval where the coefficient functions are continuous. First divide by the coefficient of \(y^{\prime}\), if necessary, to obtain this standard form.

Choose the integrating factor \(\mathrm{I}(x)=\mathrm{e}^{\int\mathrm{P}(x)\,\mathrm{d}x}\). Any one antiderivative of \(\mathrm{P}\) may be used.

\(\begin{aligned}\mathrm{I}y^{\prime}+\mathrm{I}\mathrm{P}y&=\mathrm{I}\mathrm{Q}\\\dfrac{\mathrm{d}}{\mathrm{d}x}(\mathrm{I}y)&=\mathrm{I}\mathrm{Q}\\\mathrm{I}y&=\int\mathrm{I}\mathrm{Q}\,\mathrm{d}x+C\\y&=\dfrac{1}{\mathrm{I}}\left(\int\mathrm{I}\mathrm{Q}\,\mathrm{d}x+C\right).\end{aligned}\)

Use an initial condition to determine \(C\). Check the resulting solution in the original equation and keep the interval on which that equation is defined.

The chain rule gives \(\mathrm{I}^{\prime}=\mathrm{P}\mathrm{I}\). Therefore the product rule makes the left side the derivative of \(\mathrm{I}y\):

\(\dfrac{\mathrm{d}}{\mathrm{d}x}(\mathrm{I}y)=\mathrm{I}y^{\prime}+\mathrm{I}^{\prime}y=\mathrm{I}(y^{\prime}+\mathrm{P}y)\).

To obtain the factor, solve \(\dfrac{\mathrm{I}^{\prime}}{\mathrm{I}}=\mathrm{P}\). Integration gives \(\ln|\mathrm{I}|=\int\mathrm{P}\,\mathrm{d}x+c\), so every nonzero constant multiple of \(\mathrm{e}^{\int\mathrm{P}\,\mathrm{d}x}\) is an integrating factor.

Multiplying the factor by a nonzero constant multiplies the whole transformed equation by that constant. It does not change the solution family; choose the positive exponential factor for convenience.

Differential Equations

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Integrating Factors

Integrating Factors