Answer to Question #148089 in Differential Equations for Divine Grace Jaravata

Question

Solve for the equation of the orthogonal trajectory: 1. Y = 3x - 1 + Ce-3x

Answer 1

Given: $y=3x-1+Ce^{-3x}$ ... (1)

Find the differential equation for the given family of curves by eliminating the arbitrary constant $C$ .

Differing the given equation (1) with respect to $x$ , we get

$\frac{dy}{dx}=3(1)-0+Ce^{-3x}(-3)=3-3Ce^{-3x}$

That is $\frac{dy}{dx}=3-3Ce^{-3x} ... (2)$

Eliminate $C$ from the equations (1) and (2).

From equation (2), $3Ce^{-3x} =3-\frac{dy}{dx} \Rightarrow C=e^{3x}-\frac{1}{3}e^{3x}\frac{dy}{dx}$

Substitute $C$ in equation (1), we get $y=3x-1+\left ( e^{3x}-\frac{1}{3}e^{3x}\frac{dy}{dx} \right )e^{-3x}=3x-1+1-\frac{1}{3}\frac{dy}{dx}$

$\Rightarrow y=3x-\frac{1}{3}\frac{dy}{dx}$

Replace $\frac{dy}{dx}$ by $-\frac{dx}{dy}$ to get the differential equation of the orthogonal trajectories.

Then, we get

$y=3x-\frac{1}{3}\left ( -\frac{dx}{dy} \right )=3x+\frac{1}{3}\frac{dx}{dy}$

$\frac{dx}{dy}+9x=3y$

The above is a first order linear differential equation of the form $\frac{dx}{dy}+p(y)x=g(y)$ ,

where $p(y)=9$ and $g(y)=3y$

General solution to the linear differential equation is

$x($ Integrating Factor) = $\int ($ Integrating Factor)($g(y))dy+c$ ,

where Integrating Factor = $e^{\int p(y)dy}$

Using the above, Integrating Factor = $e^{9y}$

General solution is $x(e^{9y})=\int (e^{9y})(3y)dy+c$

Use integration by parts to evaluate the right side integral.

Then, we get

$x=\frac{y}{3}-\frac{1}{27}+ce^{-9y}$

Therefore, orthogonal trajectories for the given family of curves is

$x=\frac{y}{3}-\frac{1}{27}+ce^{-9y}$ , where $c$ is arbitrary constant.