Question

Verify that when the linear differential equation $\left[\mathbf{P}\left(\mathbf{x}\right)\mathbf{y}\mathbf{-}\mathbf{Q}\left(\mathbf{x}\right)\right]\mathbf{dx}\mathbf{+}\mathbf{dy}\mathbf{=}\mathbf{0}$is multiplied by $\mathbf{\mu }\left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}$, the result is exact.

Short answer

Verified

Step-by-step solution

General form of special integrating factors

Theorem 3:

If$\frac{\left(\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\right)}{\mathbf{N}}$ is continuous and depends only on x, then$\mu \left(\mathbf{x}\right)\mathbf{=}\mathbf{exp}\left[\int \left(\frac{\left(\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\right)}{\mathbf{N}}\right)\mathbf{dx}\right]$

is an integrating factor for the equation.

If$\frac{\left(\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\right)}{\mathbf{M}}$ is continuous and depends only on y, then$\mu \left(\mathbf{y}\right)\mathbf{=}\mathbf{exp}\left[\int \left(\frac{\left(\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\right)}{\mathbf{M}}\right)\mathbf{dy}\right]$

is an integrating factor for the equation.

Evaluation method

Given that, $\left[\mathbf{P}\left(\mathbf{x}\right)\mathbf{y}\mathbf{-}\mathbf{Q}\left(\mathbf{x}\right)\right]\mathbf{dx}\mathbf{+}\mathbf{dy}\mathbf{=}\mathbf{0}······\left(1\right)$

Multiply by$\mu \left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}$on both sides. We get,

$\left[\mathbf{P}\left(\mathbf{x}\right){\mathbf{ye}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\mathbf{-}\mathbf{Q}\left(\mathbf{x}\right){\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\right]\mathbf{dx}\mathbf{+}{\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\mathbf{dy}\mathbf{=}\mathbf{0}······\left(2\right)$$\left[\mathbf{P}\left(\mathbf{x}\right){\mathbf{ye}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\mathbf{-}\mathbf{Q}\left(\mathbf{x}\right){\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\right]\mathbf{dx}\mathbf{+}{\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\mathbf{dy}\mathbf{=}\mathbf{0}······\left(2\right)$

Let $\mathbf{M}\mathbf{=}\mathbf{P}\left(\mathbf{x}\right){\mathbf{ye}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\mathbf{-}\mathbf{Q}\left(\mathbf{x}\right){\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\mathbf{,}\mathbf{N}\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}$.

Then,

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{P}\left(\mathbf{x}\right){\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{P}\left(\mathbf{x}\right){\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\end{array}$

Therefore, .

$\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}$

So, the result for this problem is exact.