Use the method discussed under “Homogeneous Equations” to solve problems 9-16. $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{t}\sqrt{{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}}{\mathbf{tx}}$

If the right-hand side of the equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$can be expressed as a function of the ratio $\frac{\mathrm{y}}{\mathrm{x}}$ alone, then we say the equation is homogeneous.

Given,$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{{\mathrm{x}}^2+\mathrm{t}\sqrt{{\mathrm{t}}^2+{\mathrm{x}}^2}}{\mathrm{tx}}$.

Evaluate it by dividing ${\mathrm{t}}^2$by both numerator and denominator.

role="math" localid="1655106866985" $\begin{array}{rcl}\frac{\mathrm{dx}}{\mathrm{dt}}& =& \frac{{\displaystyle \frac{{\mathrm{x}}^2}{{\mathrm{t}}^2}+\sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{t}}^2}}}}{{\displaystyle \raisebox{1ex}{$\mathrm{x}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{t}$}\right.}}\\ \frac{\mathrm{dx}}{\mathrm{dt}}& =& \frac{{\displaystyle {\left(\frac{\mathrm{x}}{\mathrm{t}}\right)}^2+\sqrt{1+{\left(\frac{\mathrm{x}}{\mathrm{t}}\right)}^2}}}{{\displaystyle \left(\frac{\mathrm{x}}{\mathrm{t}}\right)}}\end{array}$

Let us take $\mathrm{v}=\frac{\mathrm{y}}{\mathrm{t}}$.

Then $\mathrm{y}=\mathrm{vt}$.

By Differentiating,

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}+\mathrm{t}\frac{\mathrm{dv}}{\mathrm{dt}} \\ \frac{{\mathrm{v}}^2+\sqrt{1+{\mathrm{v}}^2}}{\mathrm{v}}=\mathrm{v}+\mathrm{t}\frac{\mathrm{dv}}{\mathrm{dt}} \end{gathered}$$

$$\begin{gathered} \mathrm{v}+\frac{\sqrt{1+{\mathrm{v}}^2}}{\mathrm{v}}=\mathrm{v}+\mathrm{t}\frac{\mathrm{dv}}{\mathrm{dt}} \\ \frac{\sqrt{1+{\mathrm{v}}^2}}{\mathrm{v}}=\mathrm{t}\frac{\mathrm{dv}}{\mathrm{dt}} \\ \frac{\sqrt{1+{\mathrm{v}}^2}}{\mathrm{v}}\frac{1}{\mathrm{dv}}=\mathrm{t}\frac{1}{\mathrm{dt}} \\ \frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^2}}\mathrm{dv}=\frac{1}{\mathrm{t}}\mathrm{dt} \end{gathered}$$

Now, integrate on both sides.

$$\begin{gathered} \int \frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^2}}\mathrm{dv}=\int \frac{1}{\mathrm{t}}\mathrm{dt} \\ \int \frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^2}}\mathrm{dv}=\ln \left|\mathrm{t}\right|+\mathrm{C} \end{gathered}$$

Integrate $\int \frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^2}}\mathrm{dv}$separately.

Let us take $\mathrm{w}=1+{\mathrm{v}}^2$.

Then, role="math" localid="1655108898121" $$\begin{gathered} \begin{array}{rcl}& & \frac{\mathrm{dw}}{\mathrm{dv}}=2\mathrm{v} \\ \mathrm{dv}=\frac{1}{2\mathrm{v}}\mathrm{dw} \\ \end{array} \end{gathered}$$

Now,

$\begin{array}{rcl}\int \frac{\mathrm{v}}{\sqrt{\mathrm{w}}}\frac{1}{2\mathrm{v}}\mathrm{dw}& =& \frac{1}{2}\int \frac{1}{\sqrt{\mathrm{w}}}\mathrm{dw}\\ & =& \frac{1}{2}\left(2\right)\sqrt{\mathrm{w}}\\ & =& \sqrt{\mathrm{w}}\end{array}$

Substitute $\mathrm{w}=1+{\mathrm{v}}^2$.

$\begin{array}{rcl}\int \frac{\mathrm{v}}{\sqrt{1+{\mathrm{v}}^2}}\mathrm{dv}& =& \sqrt{\mathrm{w}}\\ & =& \sqrt{1+{\mathrm{v}}^2}\end{array}$

Then,

$\sqrt{1+{\mathrm{v}}^2}=\ln \left|\mathrm{t}\right|+\mathrm{C}$

Substitute role="math" localid="1655108419921" $\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}}$.

$\begin{array}{rcl}\sqrt{1+{\left(\frac{\mathrm{x}}{\mathrm{t}}\right)}^2}& =& \ln \left|\mathrm{t}\right|+\mathrm{C}\\ \sqrt{1+\frac{{\mathrm{x}}^2}{{\mathrm{t}}^2}}& =& \ln \left|\mathrm{t}\right|+\mathrm{C}\end{array}$

Therefore, Homogeneous equation for the given equation is$\sqrt{\mathbf{1}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{t}}^{\mathbf{2}}}}\mathbf{=}\mathbf{ln}\mathbf{\left|}\mathbf{t}\mathbf{\right|}\mathbf{+}\mathbf{C}$.