Question

Use the method discussed under “Homogeneous Equations” to solve problems 9-16.

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{t}\sqrt{{\mathbf{t}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}}{\mathbf{tx}}$

Short answer

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

Step-by-step solution

General form of Homogeneous equation

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

Evaluate the given equation

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

Evaluate it by dividing $t^2$by both numerator and denominator.

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{{\displaystyle \frac{x^2}{t^2}}+\sqrt{1+{\displaystyle \frac{x^2}{t^2}}}}{{\displaystyle \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$t$}\right.}} \\ \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{{\left(\frac{\mathrm{x}}{\mathrm{t}}\right)}^2+\sqrt{1+{\left(\frac{\mathrm{x}}{\mathrm{t}}\right)}^2}}{\left(\frac{\mathrm{x}}{\mathrm{t}}\right)} \end{gathered}$$

Substitution method

Let us take $v=\frac{y}{x}$

Then $y=vx$

By Differentiating,

$\begin{array}{rcl}\frac{\mathbf{dx}}{\mathbf{dt}}& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{t}\frac{\mathbf{dv}}{\mathbf{dt}}\\ \left(\frac{{\mathbf{v}}^{\mathbf{2}}\mathbf{+}\sqrt{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}}{\mathbf{v}}\right)& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{t}\frac{\mathbf{dv}}{\mathbf{dt}}\\ & & \end{array}$

$\begin{array}{rcl}\left(\mathbf{v}\mathbf{+}\frac{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}}{\mathbf{v}}\right)& \mathbf{=}& \mathbf{v}\mathbf{+}\mathbf{t}\frac{\mathbf{dv}}{\mathbf{dt}}\\ \frac{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}}{\mathbf{v}}& \mathbf{=}& \mathbf{t}\frac{\mathbf{dv}}{\mathbf{dt}}\\ \frac{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}}{\mathbf{v}}\frac{\mathbf{1}}{\mathbf{dv}}& \mathbf{=}& \mathbf{t}\frac{\mathbf{1}}{\mathbf{dt}}\\ \frac{\mathbf{v}}{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{v}}^{\mathbf{2}}}}\mathbf{dv}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{t}}\mathbf{dt}\\ & & \end{array}$

Integrate the equation

Now, integrate on both sides.

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

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

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

Then, $$\begin{gathered} \frac{\mathbf{dw}}{\mathbf{dv}}\mathbf{=}\mathbf{2}\mathbf{v} \\ \mathbf{dv}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{v}}\mathbf{dw} \end{gathered}$$

Now,

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

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

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

Then,

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

Substitute $\mathbf{v}\mathbf{=}\frac{\mathbf{y}}{\mathbf{x}}$

$$\begin{gathered} \sqrt{\mathbf{1}\mathbf{+}{\left(\frac{\mathbf{x}}{\mathbf{t}}\right)}^{\mathbf{2}}}\mathbf{=}\mathbf{ln}\left|\mathbf{t}\right|\mathbf{+}\mathbf{C} \\ \sqrt{\mathbf{1}\mathbf{+}\frac{{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{t}}^{\mathbf{2}}}}\mathbf{=}\mathbf{ln}\left|\mathbf{t}\right|\mathbf{+}\mathbf{C} \end{gathered}$$

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