Advertisements
Advertisements
प्रश्न
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
उत्तर
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
संबंधित प्रश्न
Verify that \[y = ce^{tan^{- 1}} x\] is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]
Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\] is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]
Differential equation \[x\frac{dy}{dx} = 1, y\left( 1 \right) = 0\]
Function y = log x
Differential equation \[\frac{d^2 y}{d x^2} - y = 0, y \left( 0 \right) = 2, y' \left( 0 \right) = 0\] Function y = ex + e−x
x cos2 y dx = y cos2 x dy
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
Solve the following initial value problem:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]
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 which passes through the point (3, −4) and has the slope \[\frac{2y}{x}\] at any point (x, y) on it.
A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.
Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.
Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]
y2 dx + (x2 − xy + y2) dy = 0
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 |
In each of the following examples, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| y = ex | `dy/ dx= y` |
Solve the following differential equation.
`(dθ)/dt = − k (θ − θ_0)`
Solve the following differential equation.
dr + (2r)dθ= 8dθ
Choose the correct alternative.
The differential equation of y = `k_1 + k_2/x` is
Choose the correct alternative.
The integrating factor of `dy/dx - y = e^x `is ex, then its solution is
The integrating factor of the differential equation `dy/dx - y = x` is e−x.
Choose the correct alternative:
Solution of the equation `x("d"y)/("d"x)` = y log y is
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log |f(x)| + log c
∴ the general solution is
`square + log |tan y|` = log c
∴ log |tan x . tan y| = log c
`square`
This is the general solution.
Integrating factor of the differential equation `x "dy"/"dx" - y` = sinx is ______.
lf the straight lines `ax + by + p` = 0 and `x cos alpha + y sin alpha = p` are inclined at an angle π/4 and concurrent with the straight line `x sin alpha - y cos alpha` = 0, then the value of `a^2 + b^2` is
There are n students in a school. If r % among the students are 12 years or younger, which of the following expressions represents the number of students who are older than 12?
Solve the differential equation
`y (dy)/(dx) + x` = 0
Solve the differential equation
`x + y dy/dx` = x2 + y2
Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.
