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Let n be the population of the village at any time t. According to question, Now, at t=0, n = 20000 (Year 1999) Again, at t=5, n= 25000 (Year 2004) Using these values, at t =10 (Year 2009) Therefore, the population of the village in 2009 will be 31250.
Given equation is we can rewrite it as Integrate both the sides Put   put   again Put this in our equation Now, by using boundary conditions we will find the value of C It is given that  y = 0 when x = 0 at   x = 0 Now, put the value of C Therefore, the particular solution is  
Given equation is This is    type where  and  Now,                       Now, the solution of given differential equation is given by relation Now, by using boundary conditions we will find the value of C It is given that  y = 0 when  at    Now, put the value of C Therefore, the particular solution is   
Given, This is equation is in the form of    p =   and Q =  Now, I.F. =  We know that the solution of the given differential equation is:  
Given equation is Now, integrate both the sides Put Now, given equation become Now, integrate both the sides Put   again Now, by using boundary conditions we will find the value of C It is given that y = -1 when x = 0 Now, put the value of C Therefore,  the particular solution of the differential equation    is   
Given, Let  Differentiating it w.r.t. y, we get, Thus from these two equations,we get,
Given equation is we can rewrite it as Integrate both the sides Now by using boundary conditiond, we will find the value of C It is given that the curve passing through the point  So, Now, Therefore, the equation of the curve passing through the point  whose differential equation is  is   
Given equation is we can rewrite it as Now, integrate both the sides Put   Put  again Put this in our equation Now, by using boundary conditions we will find the value of C It is given that y = 1 when x = 0 Now, put the value of C Therefore,  the particular solution of the differential equation    is   
Given equation is we can rewrite it as Now, integrate on both the sides Therefore, the general solution of the differential equation    is    
Now, equation of the circle with center at (x,y) and radius r is Since, it  touch the coordinate axes in first quadrant  Therefore, x = y = r                      -(i) Differentiate it w.r.t x we will get           -(ii) Put value from equation (ii) in equation (i) Therefore,  the differential equation of the family of circles in the first quadrant which touches the coordinate axes is   
Given, Now, let y = vx Substituting the values of y and y' in the equation,  Integrating both sides we get,   Now,   Let        Now,  Let v2 = p Now, substituting the values of I1 and I2 in the above equation, we get, Thus,
Given equation is                     we can rewrite it as                    -(i) Differentiate both the sides w.r.t x                                             -(ii) Put value from equation (ii)  in (i)   Therefore, the required differential equation is                      
Given, Now, differentiating both sides w.r.t. x, Putting  values in LHS Therefore, the given function is the solution of the corresponding differential equation.
Given, Now, differentiating both sides w.r.t. x, Again, differentiating both sides w.r.t. x, Therefore, the given function is the solution of the corresponding differential equation.
Given, Now, differentiating both sides w.r.t. x, Again, differentiating both sides w.r.t. x, Therefore, the given function is the solution of the corresponding differential equation.
Given, Now, differentiating both sides w.r.t. x, Again, differentiating both sides w.r.t. x, Therefore, the given function is the solution of the corresponding differential equation.
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