Question

DY by DX into x square y cube + X Y is equal to 1

Answer 1

I hope it me help you

Answer 2

Concept: One of the two key concepts of calculus, along with integration, is differentiation. Finding a function's derivative by differentiation is one way. Finding the instantaneous rate of change in a function depending on one of its variables is a technique known as differentiation in mathematics. The most prevalent example is velocity, which is the rate at which a distance changes in relation to time. Discovering an anti-differentiation is the exact opposite of finding a derivative.

The rate of change of x with respect to y is given by dy/dx where x is one variable and y is another. It is written as f'(x) = dy/dx, where y = f(x) can be any function, and it serves as the standard expression for a function's derivative.

Given: $\frac{dy}{dx} (x^{2} y^{3} + xy) =1$

Solution:

$\frac{dy}{dx} (x^{2} y^{3} + xy) =1\\\frac{dy}{dx} = \frac{1}{x^{2} y^{3} + xy} \\\frac{dy}{dx} = x^{2} y^{3} + xy\\\frac{dy}{dx} - xy = x^{2} y^{3}\\\frac{1}{x^{2} } \frac{dy}{dx} - \frac{y}{x} = y^{3}$

substitute $\frac{1}{x} = u$

$\frac{du}{dy} = -\frac{1}{x^{2} } \frac{dx}{dy} \\-\frac{du}{dy} - uy = y^{3} \\\frac{du}{dy} + uy = -y^{3}$

I.F. =$e^{\int\limits^a_b {} \, ydy} = e^{\frac{y^2}{2} }$

$u * e^{\frac{y^2}{2} } = - \int\limits^a_b {y^3 e^{\frac{y^2}{2} } } \, dy$

$= -2(\frac{y^2}{2} -1)e^\frac{y^2}{2} +c\\= (2-y^2)e^\frac{y^2}{2} +c\\x(2-y^2)+ cxe^\frac{-y^2}{2} = 1$

Hence Proved.

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