Question

In Problems 21–26, solve the initial value problem.

$\left(\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{x}\right)\mathbf{dx}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{y}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{cosy}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{\pi }$

Short answer

The solution is $\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{sin}\mathbf{y}\mathbf{=}{\pi }^{\mathbf{2}}$.

Step-by-step solution

Evaluate the equation is exact

Here $\left(\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{x}\right)\mathbf{dx}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{y}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{cos}\mathbf{y}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\pi$

The condition for exact is $\frac{\partial \mathbf{M}}{\partial \mathbf{y}}=\frac{\partial \mathbf{N}}{\partial \mathbf{x}}$.

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\left(\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{x}\right) \\ \mathbf{N}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{y}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{cos}\mathbf{y}\mathbf{)} \end{gathered}$$

$\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{4}\mathbf{xy}\mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}$

This equation is exact.

Find the value of F(x, y)

Here

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\left(\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{x}\right) \\ \mathbf{F}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\int \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \\ \mathbf{=}\int \left(\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+}\mathbf{2}{\mathbf{y}}^{\mathbf{2}}\mathbf{x}\right)\mathbf{dx}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \\ \mathbf{=}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)} \end{gathered}$$

Determine the value of g(y)

$$\begin{gathered} \frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{N}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)} \\ \mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{y}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{y}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{cos}\mathbf{y}\mathbf{)} \\ \mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{cos}\mathbf{y} \\ \mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{sin}\mathbf{y} \end{gathered}$$

Now $\mathbf{F}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{sin}\mathbf{y}$

The solution of the differential equation is$\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{sin}\mathbf{y}\mathbf{=}\mathbf{C}$ .

Apply the initial conditions.

$$\begin{gathered} \mathbf{ln}\left|\mathbf{1}\right|\mathbf{+}{\mathbf{1}}^{\mathbf{2}}{\pi }^{\mathbf{2}}\mathbf{-}\mathbf{sin}\pi \mathbf{=}\mathbf{C} \\ \mathbf{C}\mathbf{=}{\pi }^{\mathbf{2}} \end{gathered}$$

The solution is $\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{sin}\mathbf{y}\mathbf{=}{\pi }^{\mathbf{2}}$.

Hence the solution is role="math" localid="1664179448399" $\mathbf{ln}\mathbf{}\left|x\right|\mathbf{+}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{sin}\mathbf{}\mathbf{y}\mathbf{=}{\mathit{\pi }}^{\mathbf{2}}$