Question

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Answer 1

Question:

Find the particular solution of the differential equation :

$(1 + {x}^{2}) \frac{dy}{dx} + 2xy = \frac{1}{(1 + {x}^{2}) }$

given that y = 0 if x = 1 .

Answer:

$y( 1 + {x}^{2} ) = {tan}^{ - 1}x \: - \frac{\pi}{4}$

Note:

• Particular solution : The solution of a differential equation for particular values of one or more arbitrary constants is called a particular solution of the given differential equation.

• Linear differential equations : An equation of the form ; dy/dx + Py = Q where P and Q are functions of x (or any constant) is called differential equation of the first order.

To solve these types of differential equations first we find the integrating factor of the given differential equation which is given as ;

$I.F. = {e}^{ \int Pdx}$

Then , the solution is given by ;

$y(I.F.) = \int Q(I.F.)dx + c$

where c is integration constant.

• $\int \frac{f'(x)}{f(x)}dx = log |f(x)| + c$

where c is integration constant.

• ${e}^{ log_{e}(N) } = N$

Solution:

The given differential equation is :

$= > (1 + {x}^{2}) \frac{dy}{dx} + 2xy = \frac{1}{(1 + {x}^{2}) }$

$= > \frac{dy}{dx} + \frac{2xy}{(1 + {x}^{2}) } = \frac{1}{(1 + {x}^{2} )(1 + {x}^{2}) }$

$= > \frac{dy}{dx} + \frac{2x}{(1 + {x}^{2}) }y = \frac{1}{ {(1 + {x}^{2}) }^{2} }$

Clearly ,

The above differential equation is a linear differential equation of the form :

dy/dx + Py = Q , where P = 2x/(1+x²) and

Q = 1/(1+x²)² .

Now,

The integrating factor will be given as ;

$= > I.F. = {e}^{ \int Pdx}$

$= > I.F. = {e}^{ \int \frac{2x}{1 + {x}^{2} } dx}$

$= > I.F. = {e}^{ \int log_{e}(1 + {x}^{2} ) }$

$= > I.F. = 1 + {x}^{2}$

Now,

The solution of the differential equation will be ;

$= > y(I.F.) = \int Q(I.F.)dx + c$

$= > y( 1 + {x}^{2} ) = \int (1 + {x}^{2} ) \times \frac{1}{ {(1 + {x}^{2} )}^{2} } dx$

$= > y( 1 + {x}^{2} ) = \int \frac{1}{1 + {x}^{2} } dx$

$= > y( 1 + {x}^{2} ) = {tan}^{ - 1}x \: + c \: \: ......eq-(1)$

Hence,

The eq-(1) is the general solution of the given differential equation.

Also,

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

Now,

In order to find the particular solution put x = 1 and y = 0 in eq-(1) then find the value of integration c and finally put the value of c in eq-(1).

Thus,

$= > 0.( 1 + {1}^{2} ) = {tan}^{ - 1}1 \: + c$

$= > 0 = \frac{\pi}{4} + c$

$= > c = - \frac{\pi}{4}$

Now,

Putting c = - π/4 in eq-(1) , we get ;

$= > y( 1 + {x}^{2} ) = {tan}^{ - 1}x \: - \frac{\pi}{4}$

Thus,

Above equation is the required answer.

Answer 2

Step-by-step explanation:

Prefer to the attachment

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