Find the particular solution of the differential equation "dy"/"dx" = "xy"/("x"^2+"y"^2),given that y = 1 when x = 0

Question

Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0

Sum

Solution

The given differential equation is `"dy"/"dx" = "xy"/("x"^2+"y"^2)` ....(i)

Let y = vx, then .......(ii)

`"dy"/"dx"="v" + "x" "dv"/"dx"` ....(iii)

from (i), (ii) and (iii), we get $v + x \frac{d v}{d x} = \frac{v x^{2}}{x^{2} + v^{2} x^{2}} \Rightarrow v + x \frac{d v}{d x} = \frac{v^{2}}{1 + v^{2}} \Rightarrow x \frac{d v}{d x} = \frac{v^{2} - v - v^{3}}{1 + v^{2}} \Rightarrow \frac{(1 + v^{2})}{v^{3} + v - v^{2}} d v = - \frac{d x}{x} \Rightarrow \frac{(v^{2} + 1 - v + v)}{v (v^{2} + 1 - v)} d v = - \frac{d x}{x} \Rightarrow (\frac{1}{v} + \frac{1}{v^{2} + 1 - v}) d v = - \frac{d x}{x}$

`"v" + "x" "dv"/"dx" = "vx"^2/("x"^2+"v"^2"x"^2)`

`⇒ "v" + "x" "dv"/"dx" = "v"^2/(1+"v"^2)`

`⇒ "x" "dv"/"dx" = ("v"^2-"v"-"v"^3)/(1+"v"^2)`

`⇒((1+"v"^2))/("v"^3+"v"-"v"^2)"dv" = -"dx"/"x"`

`⇒(("v"^2 +1-"v"+"v"))/("v"("v"^2+1-"v"))"dv" = -"dx"/"x"`

`⇒(1/"v" + 1/("v"^2+1-"v"))"dv" = -"dx"/"x"`

Now, Integrate both the sides $\Rightarrow \int (\frac{1}{v} + \frac{1}{v^{2} + 1 - v}) d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{v^{2} + 1 - v} d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{v^{2} - 2 \cdot v \cdot \frac{1}{2} + \frac{1}{4} + 1 - \frac{1}{4}} d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{{(v - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d v = - \int \frac{d x}{x} \Rightarrow \ln v + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 v - 1}{\sqrt{3}}) = - \ln x + c \Rightarrow \ln y + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 y - x}{\sqrt{3} x}) = c$ $\Rightarrow \int (\frac{1}{v} + \frac{1}{v^{2} + 1 - v}) d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{v^{2} + 1 - v} d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{v^{2} - 2 \cdot v \cdot \frac{1}{2} + \frac{1}{4} + 1 - \frac{1}{4}} d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{{(v - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d v = - \int \frac{d x}{x} \Rightarrow \ln v + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 v - 1}{\sqrt{3}}) = - \ln x + c \Rightarrow \ln y + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 y - x}{\sqrt{3} x}) = c$

`⇒ int (1/"v"+1/("v"^2+1-"v"))"dv" = - int "dx"/"x"`

`⇒ int1/"v""dv" + int1/("v"^2+1-"v")"dv" = - int"dx"/"x"`

`⇒int 1/"v" "dv" +int 1/("v"^2-2."v". 1/2+1/4+1-1/4)"dv" = -int"dx"/"x"`

`⇒int 1/"v" "dv" + int 1/(("v"-1/2)^2+(sqrt3/2)^2) "dv" = -int"dx"/"x"`

`⇒ "lnv" + 2/sqrt3 tan^-1((2"v"-1)/sqrt3)= - "lnx"+"c"`

`⇒"lny" + 2/sqrt3 tan^-1((2"y"-"x")/(sqrt3"x"))="c"`

It is given that y = 1 when x = 0,

Therefore c =`pi/sqrt3`

Hence, the particular solution of the given differential equation is `"ln y" + 2/sqrt3 tan^-1((2"y"-"x")/(sqrt3"x")) = pi/sqrt3.` $\ln y + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 y - x}{\sqrt{3} x}) = \frac{π}{\sqrt{3}}$

shaalaa.com

Basic Concepts of Differential Equations

Is there an error in this question or solution?

Q 19.2 Q 19.1 Q 20

2018-2019 (March) 65/4/3

APPEARS IN

2018-2019 (March) 65/4/3 (with solutions)

Q 19.2 | 4 marks

Video TutorialsVIEW ALL [2]

view
Video Tutorials For All Subjects
Basic Concepts of Differential Equations

video tutorial00:30:45
Basic Concepts of Differential Equations

video tutorial00:09:13

RELATED QUESTIONS

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

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

Differential equation	Function
\[x\frac{dy}{dx} = y\]	y = ax

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

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

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

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

Solve the following differential equation:
(xy² + 2x) dx + (x² y + 2y) dy = 0

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

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

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

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:-

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

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?

The tangent at any point (x, y) of a curve makes an angle tan⁻¹(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).

Find the equation of the curve which passes through the origin and has the slope x + 3y− 1 at any point (x, y) on it.

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.

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

The differential equation satisfied by ax² + by² = 1 is

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

x²y dx – (x³ + y³) dy = 0

Solve the differential equation sec²y tan x dy + sec²x tan y dx = 0

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

`x^2 ("d"y)/("d"x)` = x² + xy − y²

Choose the correct alternative:

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

The function y = e^x is solution ______ of differential equation

Given that `"dy"/"dx"` = ye^x and x = 0, y = e. Find the value of y when x = 1.

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

Find the particular solution of the differential equation "dy"/"dx" = "xy"/("x"^2+"y"^2),given that y = 1 when x = 0

Advertisements

Advertisements

Question

Advertisements

Solution

APPEARS IN

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

Select a course

Find the particular solution of the differential equation "dy"/"dx" = "xy"/("x"^2+"y"^2),given that y = 1 when x = 0

Advertisements

Advertisements

Question

Advertisements

SolutionShow Solution

APPEARS IN

Video TutorialsVIEW ALL [2]

RELATED QUESTIONS

Select a course

Solution