On differentiation if we get f (x,y)dy - g(x,y)dx = 0 from 2x2-3xy+y2+x+2y-8 = 0 then g(2,2)/f(1,1) =
\(\frac{11}{7}\)
-3
\(\frac{-1}{3}\)
7
To solve the problem, we start by differentiating the given expression \( 2x^2 - 3xy + y^2 + x + 2y - 8 = 0 \) implicitly with respect to \( x \).
To find \( \frac{dy}{dx} \), we differentiate each term:
\( \frac{d}{dx}(2x^2) = 4x \)
\( \frac{d}{dx}(-3xy) = -3y - 3x \frac{dy}{dx} \)
\( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \)
\( \frac{d}{dx}(x) = 1 \)
\( \frac{d}{dx}(2y) = 2 \frac{dy}{dx} \)
\( \frac{d}{dx}(-8) = 0 \)
Substituting:
\[ 4x - 3y - 3x \frac{dy}{dx} + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0 \]
Rearranging terms:
\[ (3x - 2y - 2)\frac{dy}{dx} = 4x - 3y + 1 \]
Writing in the form \( f(x,y)\,dy - g(x,y)\,dx = 0 \):
\[ (3x - 2y - 2)\,dy = (4x - 3y + 1)\,dx \]
So, \( f(x,y) = 3x - 2y - 2 \) and \( g(x,y) = 4x - 3y + 1 \).
Now,
\[ g(2,2) = 4(2) - 3(2) + 1 = 3 \]
\[ f(1,1) = 3(1) - 2(1) - 2 = -1 \]
Thus,
\[ \frac{g(2,2)}{f(1,1)} = \frac{3}{-1} = -3 \]
Hence, the required value is \( -3 \).
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A differential equation is an equation that contains one or more functions with its derivatives. The derivatives of the function define the rate of change of a function at a point. It is mainly used in fields such as physics, engineering, biology and so on.
The first-order differential equation has a degree equal to 1. All the linear equations in the form of derivatives are in the first order. It has only the first derivative such as dy/dx, where x and y are the two variables and is represented as: dy/dx = f(x, y) = y’
The equation which includes second-order derivative is the second-order differential equation. It is represented as; d/dx(dy/dx) = d2y/dx2 = f”(x) = y”.
Differential equations can be divided into several types namely