हिंदी

The Slope of the Tangent at a Point P (X, Y) on a Curve is − X Y . If the Curve Passes Through the Point (3, −4), Find the Equation of the Curve.

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
\[\frac{y^2}{2} = - \frac{x^2}{2} + \frac{25}{2}\]
\[ \Rightarrow x^2 + y^2 = 25\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 21: Differential Equations - Exercise 22.11 [पृष्ठ १३५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 21 Differential Equations
Exercise 22.11 | Q 13 | पृष्ठ १३५

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

If 1, `omega` and `omega^2` are the cube roots of unity, prove `(a + b omega + c omega^2)/(c + s omega +  b omega^2) =  omega^2`


Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.


Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].

 


Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]


Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 0, y' \left( 0 \right) = 1\] Function y = sin x


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


\[\frac{1}{x}\frac{dy}{dx} = \tan^{- 1} x, x \neq 0\]

\[\frac{dy}{dx} = x e^x - \frac{5}{2} + \cos^2 x\]

\[\frac{dy}{dx} = \frac{e^x \left( \sin^2 x + \sin 2x \right)}{y\left( 2 \log y + 1 \right)}\]

\[xy\frac{dy}{dx} = \left( x + 2 \right)\left( y + 2 \right), y\left( 1 \right) = - 1\]

Find the particular solution of edy/dx = x + 1, given that y = 3, when x = 0.


\[\frac{dy}{dx} = \left( x + y \right)^2\]

\[\left( x + y \right)^2 \frac{dy}{dx} = 1\]

\[\frac{dy}{dx} = \sec\left( x + y \right)\]

\[2xy\frac{dy}{dx} = x^2 + y^2\]

Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]


Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]


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:-

\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x} , y\left( 1 \right) = 0\]


Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]


Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]


The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by


The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is


The differential equation satisfied by ax2 + by2 = 1 is


In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-

y = ex + 1            y'' − y' = 0


Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.


Solve the differential equation:

`"x"("dy")/("dx")+"y"=3"x"^2-2`


The differential equation `y dy/dx + x = 0` represents family of ______.


Choose the correct option from the given alternatives:

The solution of `1/"x" * "dy"/"dx" = tan^-1 "x"` is


Solve the following differential equation.

`dy /dx +(x-2 y)/ (2x- y)= 0`


Solve the following differential equation.

`x^2 dy/dx = x^2 +xy - y^2`


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 the differential equation:

dr = a r dθ − θ dr


Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`


Solve the differential equation xdx + 2ydy = 0


The function y = ex is solution  ______ of differential equation


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 

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.


Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`


Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×