Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
According to the question,
\[\frac{dy}{dx} = \frac{- x}{y}\]
\[ \Rightarrow y dy = - x dx \]
ntegrating both sides with respect to x, we get
\[\int y dy = - \int x dx\]
\[ \Rightarrow \frac{y^2}{2} = - \frac{x^2}{2} + C\]
\[\text{ Since the curve passes through }\left( 3, - 4 \right),\text{ it satisfies the above equation . }\]
\[ \therefore \frac{\left( - 4 \right)^2}{2} = - \frac{3^2}{2} + C\]
\[ \Rightarrow 8 = - \frac{9}{2} + C\]
\[ \Rightarrow C = \frac{25}{2}\]
Putting the value of C, we get
\[\frac{y^2}{2} = - \frac{x^2}{2} + \frac{25}{2}\]
\[ \Rightarrow x^2 + y^2 = 25\]
APPEARS IN
संबंधित प्रश्न
Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
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
Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x
(1 + x2) dy = xy dx
xy (y + 1) dy = (x2 + 1) dx
xy dy = (y − 1) (x + 1) dx
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = \[\frac{\pi}{2}\], when x = \[\frac{\pi}{2}\]
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).
Solve the following initial value problem:-
\[y' + y = e^x , y\left( 0 \right) = \frac{1}{2}\]
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x} , y\left( 1 \right) = 0\]
The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.
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 solution of the differential equation y1 y3 = y22 is
Determine the order and degree of the following differential equations.
| Solution | D.E |
| y = aex + be−x | `(d^2y)/dx^2= 1` |
Solve the following differential equation.
`dy/dx = x^2 y + y`
Solve the following differential equation.
x2y dx − (x3 + y3) dy = 0
Choose the correct alternative.
Bacteria increases at the rate proportional to the number present. If the original number M doubles in 3 hours, then the number of bacteria will be 4M in
Solve:
(x + y) dy = a2 dx
Solve the differential equation (x2 – yx2)dy + (y2 + xy2)dx = 0
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
State whether the following statement is True or False:
The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is e–x
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
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 `"dy"/"dx" - y` = cos x is ex.
Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.
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
