Question:
The solution of $x \, dy - y \, dx + x^2 \, e^x \, dx = 0$ is :
The solution of $x \, dy - y \, dx + x^2 \, e^x \, dx = 0$ is :
Updated On: Jul 7, 2022
- $\frac{y}{x} + e^x = c $
- $\frac{x}{y} + e^x = c $
- $x + e^y = c $
- $y + e^x = c $
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The Correct Option is A
Solution and Explanation
The diff. e can be written as:
$\frac{dy}{dx} - \frac{y}{x} = - xe^{x}$
$ I.f. =e^{-\int \frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$
Thus, the solution is;
$ y \frac{1}{x} = \int -xe^{x}. \frac{1}{x} dx +c $
$ \Rightarrow \frac{y}{x} = - e^{x} +c$
$\Rightarrow \frac{y}{x} + e^{x} = c $
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.