Question

In Problems 7-12, solve the equation.

$\left(\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{x}\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Short answer

The solution for the given equation is ${\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{=}\mathbf{C}$.

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.

Iflocalid="1664267819765" $\frac{\left(\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{-}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\right)}{\mathbf{M}}$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.

Evaluate the given equation

Given,

$\left(\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{x}\right)\mathbf{dy}\mathbf{=}\mathbf{0}······\left(1\right)$

Let $\mathbf{M}\mathbf{=}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{x}$.

Then,

$\begin{array}{c}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{4}\mathbf{y}\mathbf{+}\mathbf{2}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{1}\end{array}$

Then, substitute the values to find it.

$\begin{array}{c}\frac{\left(\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}\right)}{\mathbf{N}}\mathbf{=}\frac{\mathbf{4}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{1}}{\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{x}}\\ \mathbf{=}\frac{\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{1}}{\mathbf{x}\left(\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{1}\right)}\\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\end{array}$

So, we obtain an integrating factor that is a function of x alone.

Integrating factor

Then, find the integrating factor for y alone.

Let $\mathbf{P}\left(\mathbf{x}\right)\mathbf{=}\frac{1}{\mathbf{x}}$.

Integrate on both sides.

$\begin{array}{c}\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}\mathbf{=}\int \frac{1}{\mathbf{x}}\mathbf{dx}\\ \mathbf{=}\mathbf{ln}\left|\mathbf{x}\right|\end{array}$

Now find the value of $\mu \left(\mathbf{x}\right)$.

$\begin{array}{c}\mu \left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(\mathbf{x}\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{ln}\left|\mathbf{x}\right|}\\ \mathbf{=}\mathbf{x}\end{array}$

Substitute the value of in equation (1).

$\begin{array}{c}\mathbf{x}\left(\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\right)\mathbf{dx}\mathbf{+}\mathbf{x}\left(\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{x}\right)\mathbf{dy}\mathbf{=}\mathbf{0}\\ \left(\mathbf{2}{\mathbf{xy}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{4}{\mathbf{x}}^3\right)\mathbf{dx}\mathbf{+}\left(\mathbf{2}{\mathbf{x}}^2\mathbf{y}\mathbf{+}{\mathbf{x}}^2\right)\mathbf{dy}\mathbf{=}\mathbf{0}······\left(2\right)\end{array}$

Simplifying method

Solve the equation (2).

Let $\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{3}}$.

Now integrate the value to find F.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left(\mathbf{2}{\mathbf{xy}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}\mathbf{4}{\mathbf{x}}^{\mathbf{3}}\right)\mathbf{dx}\\ \mathbf{=}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}\mathbf{g}\left(\mathbf{y}\right)\end{array}$

Differentiate F with respect to y.

$\begin{array}{c}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\\ \mathbf{=}\mathbf{N}\end{array}$

Then equalise the N values.

$\begin{array}{c}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\\ \mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}0\end{array}$

Integrate on both sides with respect to y.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(\mathbf{y}\right)\mathbf{=}\int \mathbf{0}\mathbf{dy}\\ \mathbf{g}\left(\mathbf{y}\right)\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\end{array}$

Substitute in F.

$\begin{array}{c}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{0}\\ {\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{=}\mathbf{C}\end{array}$

Hence the solution is${\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{=}\mathbf{C}$