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The solutions of the equation 3yy’ + 4x = 0 represents a:
- a)Family of circles
- b)Family of ellipses
- c)Family of Parabolas
- d)Family of hyperbolas
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The solutions of the equation 3yy’ + 4x = 0 represents a:a)Famil...
3yy’ + 4x = 0
3ydy = - 4x dx
On integration both the sides:
Where c is the constant of integration.
Thus, the solution of given differential equation represents the family of Ellipses.
Where c is the constant of integration.
Thus, the solution of given differential equation represents the family of Ellipses.
Most Upvoted Answer
The solutions of the equation 3yy’ + 4x = 0 represents a:a)Famil...
Family of Ellipses
The given differential equation 3yy' + 4x = 0 can be rearranged to give y' = -4x / 3y. This is in the standard form of a first-order differential equation. To find the solutions of this equation, we need to integrate it with respect to x.
Integrating the Equation
By integrating the equation y' = -4x / 3y with respect to x, we get:
∫(1/y) dy = ∫(-4x/3) dx
This simplifies to:
ln|y| = -2x^2 + C
Where C is the constant of integration.
Exponential Form
Taking the exponential of both sides, we get:
|y| = e^(-2x^2 + C)
|y| = e^C * e^(-2x^2)
|y| = Ae^(-2x^2)
Where A = e^C is a non-zero constant.
Ellipse Equation
The equation obtained, |y| = Ae^(-2x^2), represents a family of ellipses. This is because the general form of an ellipse centered at the origin is given by:
(x/a)^2 + (y/b)^2 = 1
Comparing this with the equation obtained, we see that it is of the form:
y = Be^(Ax^2)
Which represents a family of ellipses with foci on the y-axis.
Therefore, the solutions of the given differential equation represent a family of ellipses.
The given differential equation 3yy' + 4x = 0 can be rearranged to give y' = -4x / 3y. This is in the standard form of a first-order differential equation. To find the solutions of this equation, we need to integrate it with respect to x.
Integrating the Equation
By integrating the equation y' = -4x / 3y with respect to x, we get:
∫(1/y) dy = ∫(-4x/3) dx
This simplifies to:
ln|y| = -2x^2 + C
Where C is the constant of integration.
Exponential Form
Taking the exponential of both sides, we get:
|y| = e^(-2x^2 + C)
|y| = e^C * e^(-2x^2)
|y| = Ae^(-2x^2)
Where A = e^C is a non-zero constant.
Ellipse Equation
The equation obtained, |y| = Ae^(-2x^2), represents a family of ellipses. This is because the general form of an ellipse centered at the origin is given by:
(x/a)^2 + (y/b)^2 = 1
Comparing this with the equation obtained, we see that it is of the form:
y = Be^(Ax^2)
Which represents a family of ellipses with foci on the y-axis.
Therefore, the solutions of the given differential equation represent a family of ellipses.
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The solutions of the equation 3yy’ + 4x = 0 represents a:a)Famil...
3yy’ + 4x = 0
3ydy = - 4x dx
On integration both the sides:
Where c is the constant of integration.
Thus, the solution of given differential equation represents the family of Ellipses.
Where c is the constant of integration.
Thus, the solution of given differential equation represents the family of Ellipses.
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The solutions of the equation 3yy’ + 4x = 0 represents a:a)Family of circlesb)Family of ellipsesc)Family of Parabolasd)Family of hyperbolasCorrect answer is option 'B'. Can you explain this answer?
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