Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b: \(\frac{x}{a}+\frac{y}{b}=1\)
Solution and Explanation
\(\frac{x}{a}+\frac{y}{b}=1\)
Differentiating both sides of the given equation with respect to x, we get:
\(\frac{1}{a}+\frac{1}{b}\frac{dy}{dx}=0\)
\(\frac{1}{a}+\frac{1}{b}y=0\)
Again, differentiating both sides with respect to x, we get:
\(0+\frac{1}{b}y=0\)
\(\frac{1}{b}y=0\)
\(y=0\)
Hence,the required differential equation of the given curve is y=0.
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Determine whether each of the following relations are reflexive, symmetric, and transitive.
- Relation R in the set A={1,2,3,...13,14} defined as R={(x,y): 3x-y=0}.
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Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
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Concepts Used:
General Solutions to Differential Equations
A relation between involved variables, which satisfy the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called the general solution and the solution free from arbitrary constants is called particular solution.
For example,
Read More: Formation of a Differential Equation