Question: In Problems 1 - 30, solve the equation.

$\left[\mathbf{1}\mathbf{+}{\left(\mathbf{1}\mathbf{+}{\mathbf{x}}^2\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}{\mathbf{y}}^2\right)}^{-1}\right]\mathbf{dx}\mathbf{+}\left[{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}{\left(\mathbf{1}\mathbf{+}{\mathbf{x}}^2\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}{\mathbf{y}}^2\right)}^{-1}\right]\mathbf{dy}\mathbf{=}\mathbf{0}$

Definition of Initial Value Problem:By an initial value problem for an nth-order differential equation$\mathbf{F}\left(\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{,}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{,}\mathrm{...}\mathbf{,}\frac{{\mathbf{d}}^n\mathbf{y}}{{\mathbf{dx}}^n}\right)\mathbf{=}\mathbf{0}$ we mean: Find a solution to the differential equation on an interval I that satisfies x₀ at the n initial conditions

$\begin{array}{c}\mathbf{y}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_0\mathbf{,}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_1\mathbf{,}\\ .\\ .\\ .\\ \frac{{\mathbf{d}}^{n-1}\mathbf{y}}{{\mathbf{dx}}^{n-1}}\left({\mathbf{x}}_0\right)\mathbf{=}{\mathbf{y}}_{n-1}\mathbf{,}\mathbf{,}\end{array}$

Where${\mathbf{x}}_0\in \mathbf{I}$ and${\mathbf{y}}_0\mathbf{,}{\mathbf{y}}_1\mathbf{,}\mathrm{...}\mathbf{,}{\mathbf{y}}_{n-1}$ are given constants.

Formulae to be used:

• Integration by parts:.$\int \mathbf{udv}\mathbf{=}\mathbf{uv}\mathbf{-}\int \mathbf{vdu}$
• $\int {\mathbf{x}}^a\mathbf{dx}\mathbf{=}\frac{{\mathbf{x}}^{a+1}}{a+1}\mathbf{+}\mathbf{C}$.
• $\int \frac{1}{x}\mathbf{dx}\mathbf{=}\mathbf{ln}\left|x\right|\mathbf{+}\mathbf{C}$.

Given that,$\left[\mathbf{1}\mathbf{+}{\left(\mathbf{1}\mathbf{+}{\mathbf{x}}^2\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}{\mathbf{y}}^2\right)}^{-1}\right]\mathbf{dx}\mathbf{+}\left[{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}\mathbf{+}{\left(\mathbf{1}\mathbf{+}{\mathbf{x}}^2\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}{\mathbf{y}}^2\right)}^{-1}\right]\mathbf{dy}\mathbf{=}\mathbf{0}\mathrm{......}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Evaluate it.

$\begin{array}{c}\left[\mathbf{1}\mathbf{+}{\left(\mathbf{1}\mathbf{+}{\mathbf{x}}^2\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}{\mathbf{y}}^2\right)}^{-1}\right]\mathbf{dx}\mathbf{+}\left[{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}\mathbf{+}{\left(\mathbf{1}\mathbf{+}{\mathbf{x}}^2\mathbf{+}\mathbf{2}\mathbf{xy}\mathbf{+}{\mathbf{y}}^2\right)}^{-1}\right]\mathbf{dy}\mathbf{=}\mathbf{0}\\ \left[\mathbf{1}\mathbf{+}{\left(\mathbf{1}\mathbf{+}{\left(x+y\right)}^2\right)}^{-1}\right]\mathbf{dx}\mathbf{+}\left[{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}\mathbf{+}{\left(\mathbf{1}\mathbf{+}{\left(x+y\right)}^2\right)}^{-1}\right]\mathbf{dy}\mathbf{=}\mathbf{0}\\ \left[\mathbf{1}\mathbf{+}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^2}\right]\mathbf{dx}\mathbf{+}\left[{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}\mathbf{+}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^2}\right]\mathbf{dy}\mathbf{=}\mathbf{0}\end{array}$

Let us check whether the given equation is exact or not.

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

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{\mathbf{2}\left(x+y\right)}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^2}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\frac{\mathbf{2}\left(x+y\right)}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^2}\end{array}$

So, the given equation is exact.

Now, let us assume $\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{+}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^2}$.

Integrate on both sides.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left(\mathbf{1}\mathbf{+}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^2}\right)\mathbf{dx}\\ \mathbf{=}{\mathbf{tan}}^{-1}\left(x+y\right)\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{g}\left(y\right)\end{array}$

Differentiate the F with respect to y.

$\begin{array}{c}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^2}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\\ =N\end{array}$

Equalise the values of N.

$\begin{array}{c}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^2}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}\mathbf{+}\frac{1}{\mathbf{1}\mathbf{+}{\left(x+y\right)}^2}\\ \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}{\mathbf{y}}^{\mathbf{-}\frac{1}{2}}\end{array}$

Integrate on both sides.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\int {\mathbf{y}}^{\mathbf{-}\frac{1}{2}}\mathbf{dy}\\ \mathbf{g}\left(y\right)\mathbf{=}2{\mathbf{y}}^{\frac{1}{2}}\mathbf{+}{\mathbf{C}}_1\end{array}$

Substitute in equation of F.

${\mathbf{tan}}^{-1}\left(x+y\right)\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{2}\sqrt{y}\mathbf{=}\mathbf{C}$

Hence, the solution of the given initial value problem is ${\mathbf{tan}}^{-1}\left(x+y\right)\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{2}\sqrt{y}\mathbf{=}\mathbf{C}$.