Advertisements
Advertisements
Question
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
Advertisements
Solution
The equation of the parabola having vertex at origin and axis along the positive direction of x-axis is given by
y2 =4ax .....(1)
Since there is only one parameter, so we differentiate it only once.
Differentiating with respect to x, we get
\[2y\frac{dy}{dx} = 4a\]
Substituting the value of 4a in (1), we get
\[y^2 = 2y\frac{dy}{dx} \times x\]
\[ \Rightarrow y^2 = 2xy\frac{dy}{dx}\]
\[ \Rightarrow y^2 - 2xy\frac{dy}{dx} = 0\]
\[\]
APPEARS IN
RELATED QUESTIONS
Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].
Show that Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x + y\frac{dy}{dx} = 0\]
|
\[y = \pm \sqrt{a^2 - x^2}\]
|
Differential equation \[\frac{dy}{dx} = y, y\left( 0 \right) = 1\]
Function y = ex
(y2 + 1) dx − (x2 + 1) dy = 0
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
If y(x) is a solution of the different equation \[\left( \frac{2 + \sin x}{1 + y} \right)\frac{dy}{dx} = - \cos x\] and y(0) = 1, then find the value of y(π/2).
y ex/y dx = (xex/y + y) dy
Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + 2y = e^{- 2x} \sin x, y\left( 0 \right) = 0\]
Solve the following initial value problem:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
The slope of the tangent at a point P (x, y) on a curve is \[\frac{- x}{y}\]. If the curve passes through the point (3, −4), find the equation of the curve.
Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).
Which of the following differential equations has y = C1 ex + C2 e−x as the general solution?
Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .
Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Find the equation of the plane passing through the point (1, -2, 1) and perpendicular to the line joining the points A(3, 2, 1) and B(1, 4, 2).
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| xy = log y + k | y' (1 - xy) = y2 |
Solve the following differential equation.
`dy/dx + y = e ^-x`
Solve the following differential equation.
`dy/dx + y` = 3
Choose the correct alternative.
The differential equation of y = `k_1 + k_2/x` is
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
The solution of `dy/ dx` = 1 is ______
State whether the following is True or False:
The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.
For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0
Solve the following differential equation
`x^2 ("d"y)/("d"x)` = x2 + xy − y2
A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution
Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0
The differential equation of all non horizontal lines in a plane is `("d"^2x)/("d"y^2)` = 0
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
