हिंदी

3x2 Dy = (3xy + Y2) Dx - Mathematics

Advertisements
Advertisements

प्रश्न

3x2 dy = (3xy + y2) dx

Advertisements

उत्तर

We have, 
\[3 x^2 dy = \left( 3xy + y^2 \right) dx\]
\[ \Rightarrow \frac{dy}{dx} = \frac{3xy + y^2}{3 x^2}\]
This is a homogeneous differential equation .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}, \text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{3v x^2 + v^2 x^2}{3 x^2}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{3v + v^2}{3}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{v^2}{3}\]
\[ \Rightarrow \frac{3}{v^2}dv = \frac{1}{x}dx\]
Integrating both sides, we get
\[3\int\frac{1}{v^2}dv = \int\frac{1}{x}dx\]
\[ \Rightarrow - 3 \times \frac{1}{v} = \log \left| x \right| + C\]
\[ \Rightarrow - \frac{3}{v} = \log \left| x \right| + C\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow \frac{- 3x}{y} = \log \left| x \right| + C\]
\[\text{ Hence, }\frac{- 3x}{y} = \log \left| x \right| + C\text{ is the required solution }.\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 22: Differential Equations - Exercise 22.09 [पृष्ठ ८३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 22 Differential Equations
Exercise 22.09 | Q 14 | पृष्ठ ८३

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

\[\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 + xy = 0\]

\[x^2 \left( \frac{d^2 y}{d x^2} \right)^3 + y \left( \frac{dy}{dx} \right)^4 + y^4 = 0\]

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


Show that y = e−x + ax + b is solution of the differential equation\[e^x \frac{d^2 y}{d x^2} = 1\]

 


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


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


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


\[\frac{dy}{dx} = \cos^3 x \sin^2 x + x\sqrt{2x + 1}\]

\[\frac{dy}{dx} = \frac{1 - \cos 2y}{1 + \cos 2y}\]

\[y\sqrt{1 + x^2} + x\sqrt{1 + y^2}\frac{dy}{dx} = 0\]

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

(y + xy) dx + (x − xy2) dy = 0


\[\left( x - 1 \right)\frac{dy}{dx} = 2 x^3 y\]

Solve the following differential equation:
\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]


2xy dx + (x2 + 2y2) dy = 0


Solve the following initial value problem:-

\[\frac{dy}{dx} + 2y \tan x = \sin x; y = 0\text{ when }x = \frac{\pi}{3}\]


The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?


The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.


Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?


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 passing through the point \[\left( 1, \frac{\pi}{4} \right)\]  and tangent at any point of which makes an angle tan−1  \[\left( \frac{y}{x} - \cos^2 \frac{y}{x} \right)\] with x-axis.


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.


y2 dx + (x2 − xy + y2) dy = 0


Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]


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

Determine the order and degree of the following differential equations.

Solution D.E.
y = 1 − logx `x^2(d^2y)/dx^2 = 1`

Solve the following differential equation.

`dy/dx + y = e ^-x`


The solution of `dy/dx + x^2/y^2 = 0` is ______


Solve the differential equation:

`e^(dy/dx) = x`


Solve the following differential equation y log y = `(log  y - x) ("d"y)/("d"x)`


Choose the correct alternative:

Solution of the equation `x("d"y)/("d"x)` = y log y is


The solution of differential equation `x^2 ("d"^2y)/("d"x^2)` = 1 is ______


Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)

Solution: `("d"y)/("d"x)` = cos(x + y)    ......(1)

Put `square`

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

∴ `square` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int square  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

∴ `square` = x + c


Find the particular solution of the following differential equation

`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.

Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x

∴ `1/"e"^(2y)  "d"y` = cos x dx

Integrating, we get

`int square  "d"y` = cos x dx

∴ `("e"^(-2y))/(-2)` = sin x + c1

∴ e–2y = – 2sin x – 2c1

∴ `square` = c, where c = – 2c

This is general solution.

When x = `pi/6`, y = 0, we have

`"e"^0 + 2sin  pi/6` = c

∴ c = `square`

∴ particular solution is `square`


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


Integrating factor of the differential equation `x "dy"/"dx" - y` = sinx is ______.


Integrating factor of the differential equation `"dy"/"dx" - y` = cos x is ex.


Solve: `("d"y)/("d"x) = cos(x + y) + sin(x + y)`. [Hint: Substitute x + y = z]


The differential equation of all non horizontal lines in a plane is `("d"^2x)/("d"y^2)` = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×