मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Solve the following differential equation x2 dydx = x2 + xy − y2 - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Solve the following differential equation

`x^2  ("d"y)/("d"x)` = x2 + xy − y2 

बेरीज

उत्तर

`x^2  ("d"y)/("d"x)` = x2 + xy − y2 

∴ `("d"y)/("d"x) = 1 + y/x - (y/x)^2`   .....(i)

Put `y/x` = t      .....(ii)

∴ y = tx

Differentiating w.r.t. x, we get

`("d"y)/("d"x) = "t" + x  ("dt")/("d"x)`   .....(iii)

Substituting (ii) and (iii) in (i), we get

`"t" + x  "dt"/("d"x)` = 1 + t − t2

∴ `x  "dt"/("d"x)` = 1 − t2

∴ `"dt"/(1 - "t"^2) = ("d"x)/x`

Integrating on both sides, we get

`int  "dt"/(1 - "t"^2) = int ("d"x)/x`

∴ `1/2 log|(1 + t)/(1 - t)|` = log |x| + log |c1|

∴ `log |(1 + y/x)/(1 - y/x)|` = 2log |x| + 2log |c1|

∴ `log|(x + y)/(x - y)|` = log |x2| + log |c12|

∴ `log|(x + y)/(x - y)|` = log |c1x2|

∴ `(x + y)/(x - y)` = c12x2

∴ `(x + y)/(x - y)` = cx2, where c = c12 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1.8: Differential Equation and Applications - Q.5

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`


\[y\frac{d^2 x}{d y^2} = y^2 + 1\]

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

Verify that y = \[\frac{a}{x} + b\] is a solution of the differential equation
\[\frac{d^2 y}{d x^2} + \frac{2}{x}\left( \frac{dy}{dx} \right) = 0\]


Verify that y2 = 4ax is a solution of the differential equation y = x \[\frac{dy}{dx} + a\frac{dx}{dy}\]


Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]


Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\]  satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]


Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\]  is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]


For the following differential equation verify that the accompanying function is a solution:

Differential equation Function
\[x + y\frac{dy}{dx} = 0\]
\[y = \pm \sqrt{a^2 - x^2}\]

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 \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} = 0, y \left( 0 \right) = 2, y'\left( 0 \right) = 1\]

Function y = ex + 1


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

\[\left( x + 2 \right)\frac{dy}{dx} = x^2 + 3x + 7\]

\[\frac{dy}{dx} = \log x\]

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

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

C' (x) = 2 + 0.15 x ; C(0) = 100


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

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

\[\frac{dy}{dx} = \frac{1 + y^2}{y^3}\]

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

x cos y dy = (xex log x + ex) dx


\[\frac{dy}{dx} = \frac{x e^x \log x + e^x}{x \cos y}\]

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

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

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

(1 − x2) dy + xy dx = xy2 dx


(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0


y (1 + ex) dy = (y + 1) ex dx


\[\frac{dy}{dx} = 1 - x + y - xy\]

(y2 + 1) dx − (x2 + 1) dy = 0


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

\[\frac{dy}{dx} = y \tan 2x, y\left( 0 \right) = 2\] 

\[2x\frac{dy}{dx} = 5y, y\left( 1 \right) = 1\]

\[\frac{dy}{dx} = 2 e^{2x} y^2 , y\left( 0 \right) = - 1\]

\[\frac{dy}{dx} = 1 + x^2 + y^2 + x^2 y^2 , y\left( 0 \right) = 1\]

Solve the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\], given that y = 1, when x = 0.


Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.


In a bank principal increases at the rate of r% per year. Find the value of r if ₹100 double itself in 10 years (loge 2 = 0.6931).


In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.


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

\[\frac{dy}{dx} = \frac{y - x}{y + x}\]

\[\frac{dy}{dx} = \frac{x + y}{x - y}\]

y ex/y dx = (xex/y + y) dy


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


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

 

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:-

\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]


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 \[x = \frac{\pi}{2}\]


In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]


The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.


The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is


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


The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when


Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]


What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?


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 ab are arbitrary constant.


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


If a + ib = `("x" + "iy")/("x" - "iy"),` prove that `"a"^2 +"b"^2 = 1` and `"b"/"a" = (2"xy")/("x"^2 - "y"^2)`


Solve the differential equation:

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


Choose the correct option from the given alternatives:

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


Determine the order and degree of the following differential equations.

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

Find the differential equation whose general solution is

x3 + y3 = 35ax.


Solve the following differential equation.

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


For the following differential equation find the particular solution.

`(x + 1) dy/dx − 1 = 2e^(−y)`,

when y = 0, x = 1


Solve the following differential equation.

xdx + 2y dx = 0


Solve the following differential equation.

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


Solve the following differential equation.

(x2 − y2 ) dx + 2xy dy = 0


Solve the following differential equation.

`xy  dy/dx = x^2 + 2y^2`


Solve the following differential equation.

`(x + a) dy/dx = – y + a`


Choose the correct alternative.

The solution of `x dy/dx = y` log y is


A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.


x2y dx – (x3 + y3) dy = 0


y dx – x dy + log x dx = 0


Select and write the correct alternative from the given option for the question

Bacterial increases at the rate proportional to the number present. If original number M doubles in 3 hours, then number of bacteria will be 4M in


Solve the differential equation `("d"y)/("d"x) + y` = e−x 


Solve the differential equation xdx + 2ydy = 0


Solve the following differential equation `("d"y)/("d"x)` = x2y + y


For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0


Solve the following differential equation

`yx ("d"y)/("d"x)` = x2 + 2y2 


Solve the following differential equation y2dx + (xy + x2) dy = 0


Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0

y = `a + b/x`

`(dy)/(dx) = square`

`(d^2y)/(dx^2) = square`

Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`

= `x square + 2 square`

= `square`

Hence y = `a + b/x` is solution of `square`


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.


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`


An appropriate substitution to solve the differential equation `"dx"/"dy" = (x^2 log(x/y) - x^2)/(xy log(x/y))` is ______.


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


Solve: ydx – xdy = x2ydx.


The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×