If the curve y=f(x) passing through the point (1,2) and...

If the curve $y=f(x)$ passing through the point $(1,2)$ and satisfies the differential equation $x d y+\left(y-x^{3} y^{2}\right)$ $d x=0$, then
(A) $x y=\frac{1}{2}$
(B) $x^{3} y=2$
(C) $\frac{1}{x y}=2$
(D) None of these

Key Concepts

First-order Differential Equations

This concept involves equations that include the first derivative of an unknown function with respect to an independent variable. Techniques for solving these equations include methods such as separation of variables, integrating factors, and substitution, all of which are essential tools in understanding the behavior of dynamical systems modeled by such equations.

Separable Equations

Separable equations are a class of first-order differential equations where the variables can be separated on opposite sides of the equation. This method involves expressing the equation in the form f(y) dy = g(x) dx, allowing both sides to be integrated independently, leading to implicit or explicit solutions for the unknown function.

Variable Substitution

This technique is used to simplify differential equations by introducing a new variable that reduces the equation to a more tractable form. In many cases, an appropriate substitution can transform a non-separable equation into a separable one or reveal underlying structures that facilitate integration and solution of the differential equation.

Initial Value Problems

An initial value problem (IVP) specifies the value of the function at a certain point along with the differential equation. This additional information is crucial for determining a unique solution among the family of possible solutions, ensuring that the curve or solution curve satisfies both the differential equation and the initial condition.

Transcript

00:01 We have this curve y equals your 4 of x which passes through the point 1 comma 2 and the differential equation given is x -di -y plus y plus x cube y squared d x equals 0 and we have to choose one of the given options so basically we have to solve this differential equation let me rewrite this equation as like this that is x dy i do the distribution of the second term i'll be getting y d x d x and then and then when i multiply this x cube y squared with d x, i'll be getting x cube y squared dx, which i transpose to the right side.

00:38 When i do that, i'll be getting x cube y squared dx on the right side.

00:44 Now i divide both sides by this term x squared y squared.

00:49 So therefore, i'll be getting x, dy, plus y, dx, divided by x squared, y squared on the left side.

00:57 And when i divide this right side term, i'll just be getting negative x dx on the right side.

01:04 Now, if you could look at the terms on the left side, in its numerator, we have this term x, d y, and this one is x squared y squared.

01:15 So i can rewrite the numerator as basically the derivative of x, y, and also this x squared y squared, i can rewrite this as xy when it is squared.

01:25 So this is equal to negative x dx.

01:29 Now we have separated the variables.

01:30 We can integrate both sides.

01:32 So i'm going to do that.

01:34 So integrate both sides.

01:36 Observe that on the left side, we have this derivative of something like dx over x squared.

01:43 So you can apply this standard integration formula.

01:47 This can be written as negative of x squared dx.

01:50 And then when you apply the power rule, we'll be getting negative of...