Advertisements
Advertisements
Question
The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).
Advertisements
Solution
According to the question,
\[\frac{dy}{dx} = x^2\]
\[\Rightarrow dy = x^2 dx\]
Integrating both sides with respect to x, we get
\[\int dy = \int x^2 dx\]
\[ \Rightarrow y = \frac{x^3}{3} + C\]
\[\text{ Since the curve passes through }\left( - 1, 1 \right), \text{ it satisfies the above equation .} \]
\[ \therefore 1 = \frac{- 1}{3} + C\]
\[ \Rightarrow C = 1 + \frac{1}{3}\]
\[ \Rightarrow C = \frac{4}{3}\]
Putting the value of C, we get
\[y = \frac{x^3}{3} + \frac{4}{3}\]
\[ \Rightarrow 3y = x^3 + 4\]
APPEARS IN
RELATED QUESTIONS
Show that the differential equation of which y = 2(x2 − 1) + \[c e^{- x^2}\] is a solution, is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].
Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 1\] Function y = sin x + cos x
(1 + x2) dy = xy dx
y (1 + ex) dy = (y + 1) ex dx
Solve the following differential equation:
(xy2 + 2x) dx + (x2 y + 2y) dy = 0
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
2xy dx + (x2 + 2y2) dy = 0
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x} , y\left( 1 \right) = 0\]
Solve the following initial value problem:
\[x\frac{dy}{dx} + y = x \cos x + \sin x, y\left( \frac{\pi}{2} \right) = 1\]
Find the equation of the curve which passes through the point (1, 2) and the distance between the foot of the ordinate of the point of contact and the point of intersection of the tangent with x-axis is twice the abscissa of the point of contact.
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.
Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.
The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by
The equation of the curve whose slope is given by \[\frac{dy}{dx} = \frac{2y}{x}; x > 0, y > 0\] and which passes through the point (1, 1) is
Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .
Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constant.
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
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).
Choose the correct option from the given alternatives:
The solution of `1/"x" * "dy"/"dx" = tan^-1 "x"` is
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 |
Find the differential equation whose general solution is
x3 + y3 = 35ax.
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 0
Solve the following differential equation.
`dy/dx + y = e ^-x`
Solve the differential equation:
`e^(dy/dx) = x`
Solve:
(x + y) dy = a2 dx
Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`
Solve the following differential equation y2dx + (xy + x2) dy = 0
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
Why is the equation \[x\frac{dy}{dx} + y = 0\] classified as a differential equation?
