How do you solve this differential equation #dy/dx=(-x)/y# when #y=3# and #x=4# ?

Question

Charles Anctil · Accepted Answer

# y^2 = 25 - x^2 #

We have: # dy/dx=(-x)/y# with #y=3# when #x=4# Since this ODE can be separated, we can write: # y \ dy/dx = -x # Afterward, we are able to "separate the variables": # int \ y \ dy = - \ int \ x \ dx # Afterward, we can easily integrate to obtain: # 1/2y^2 = - 1/2x^2 + C # Given the initial condition #y(4)=3# then: # 1/2 * 9 = - 1/2 * 16 + C => C = 25/2 # Thus, the Specific Resolution is: # 1/2y^2 = - 1/2x^2 + 25/2 # # :. y^2 = 25 - x^2 # Which we note is a circle of radius #5# centred on the origin.

Isaiah Allen · Answer

undefined To solve the differential equation $\frac{dy}{dx} = -\frac{x}{y}$ given that $y = 3$ when $x = 4$, we can use separation of variables.

Separating variables:
$$\frac{dy}{y} = -\frac{dx}{x}$$

Integrating both sides:
$$\int \frac{1}{y} \, dy = -\int \frac{1}{x} \, dx$$

$$\ln|y| = -\ln|x| + C$$

Where $C$ is the constant of integration.

Rewriting in exponential form:
$$|y| = \frac{1}{|x|}e^C$$

Given that $y = 3$ when $x = 4$, we can substitute these values into the equation and solve for $C$:
$$3 = \frac{1}{4}e^C$$
$$12 = e^C$$

Taking the natural logarithm of both sides:
$$\ln(12) = C$$

So the equation becomes:
$$|y| = \frac{1}{|x|} \cdot 12$$

$$y = \pm \frac{12}{x}$$

Thus, the general solution to the differential equation is $y = \pm \frac{12}{x}$.

Ezra Adams · Answer

undefined To solve the given differential equation $ \frac{dy}{dx} = -\frac{x}{y} $ when $ y = 3 $ and $ x = 4 $, we'll follow these steps:

1. Substitute $ y = 3 $ and $ x = 4 $ into the equation to find the value of $ \frac{dy}{dx} $.
2. Solve for $ \frac{dy}{dx} $ at $ y = 3 $ and $ x = 4 $.

$ \frac{dy}{dx} = -\frac{x}{y} $

Substitute $ x = 4 $ and $ y = 3 $:

$ \frac{dy}{dx} = -\frac{4}{3} $

Therefore, $ \frac{dy}{dx} = -\frac{4}{3} $ when $ y = 3 $ and $ x = 4 $.