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Differential Equations Test - 8
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Differential Equations Test - 8
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  • Question 1/10
    1 / -0.25

    The solution of the differential equation   is :

    Solutions

     dy/dx = y/x - (y2 /x2 + 1)½ …………………(1)
    the equation is homogenous,so y = Vx
    dy/dx = V + xdv/dx  
    Put the value of dy/dx in eq(1)
    V + xdv/dx = V - (V2 + 1)½
    = ∫-dv(v2 + 1)½ = ∫dx/x
    = log|v + (v2 + 1)½ | = -logx + logc
    = log|y/x + ((y/x)2 + 1)½ | =  log c/x  
    =|y/x + ((y/x)2 + 1)½ | =  c/x

  • Question 2/10
    1 / -0.25

    The solution of the differential equation is :

    Solutions

    dy/dx = (x2 + 3y2 )/2xy ………….(1)
    Let y = vx
    dy/dx = v + xdv/dx
    Substitute the value of y and dy/dx in (1)
    v + x dv/dx = (1+3v2 )/2v
    x dv/dx = (1+3v2 )/2v - v
    x dv/dx = (1 + 3v2 - 2v2 )/2
    x dv/dx = (1+ v2 )/2v
    2v/(1+v2 ) dv = dx/x …………(2)
    Integrating both the sides
    ∫2v/(1+v2 ) dv = ∫dx/x
    Put t = 1 + v2
    dt = 2vdv
    ∫dt/t  = ∫dx/x
    =>log|t| = log|x| + log|c|
    =>log|t/x| = log|c|
    t/x = +- c
    (1+v2 )/x = +-c
    (1 + (y2 )/(x2 ))/x = +-c
    x2 + y2 = Cx3 ……….(3)
    y(1) = 3
    1 + 9 = c(1)3
    c = 10
    From eq(3), we get x2 + y2 = 10x3

  • Question 3/10
    1 / -0.25

    The first order, first degree differential equation y ’= f(x,y)  is said to be homogeneous, if ​

    Solutions

  • Question 4/10
    1 / -0.25

    The solution of the differential equation   is :

    Solutions

    Given x dy/dx = y(log y - log x+1)
    logy-logx=log(y/x)
    dy/dx=y/x(log(y/x) +1)
    substituting y=vx
    dy/dx=v+xdv/dx
    =>dv/(vlogv)=dx/x, integrating we get
    log(logv)=log(cx)
    log(y/x) = cx
    =>y/x = e(cx)
    =>y = xe(cx)

  • Question 5/10
    1 / -0.25

    The solution of the differential equation   is : 

    Solutions

    This is clearly a Homogenous differential equation, as RHS is expressed only in terms of y/x.
    To Solve this, Lety/x=t
    ⟹y=xt
    ⟹dy/dx = t+xdt/dx
    By (i)
    ⟹tant+t = t+xdt/dx
    ⟹tant=xdt/dx
    ⟹dx/x=dt/tant
    Integrating both sides,
    ⟹∫dx/x = ∫cot tdt
    ⟹logx = log(sint)+logC
    ⟹logx=log(C siny/x)
    ⟹x = Csin(y/x)

  • Question 6/10
    1 / -0.25

    The solution of the differential equationis :

    Solutions

     x -(xy)½ dy = ydx
    [x-(xy)½ ] - y = dx/dy
    dx/dy = x/y - [(xy)½ ]/y ……………….(1)
    Let V = x/y
    x = Vy
    dx/dy = V + ydv/dy ……..(2)
    V + ydv/dy = V - (V)½
    ydv/dy = -(V)½
    dv/(V)½  = -dy/y
    Integrating both the sides, we get
    (V(-½+1 ))/(-½+ 1) = -log y + c
    2(V)½ = -log y + c
    2(y/x)½ = -log y + c
    2(x)½ = (y)½ (-log y + c)
    2(x)½  = (y)1 /2log y + c(y)½  
    (x)½ = [(y)½ ]/2 log y + [c(y)½ ]/2

  • Question 7/10
    1 / -0.25

    The solution of the differential equation is :

    Solutions

    y/x cos y/x dx −(x/y sin y/x + cosy/x) dy = 0
    ⇒dy/dx = (y/x cos y/x)/(x/y sin y/x + cos y/x)
    This is a homogeneous differential equation.
    Putting y = vx and dy/dx = v+xdv/dx, we have v+xdv/dx = (v2 cosv)/(sinv+vcosv)
    ⇒xdv/dx = (v2 cosv)/(sinv+vcosv) - v
    ⇒x dv/dx = (v2 cosv - v2 cosv - vsinv )/(sinv+vcosv) 
    ⇒x dv/dx = - [v sinv/(sinv + vcosv)]
    ⇒∫[(sinv + vcosv)/v sinv]dv = ∫dx/x
    Integrating both the sides, we get
    ∫(cot v + 1/v)dv = - ln(x) + c
    ln(sin v) + ln(v) = -ln(x) + c
    ln(sin(y/x) + ln(y/x) + ln(x) = c
    ln(y/x sin y/x * x) + c
    y sin(y/x) = c

  • Question 8/10
    1 / -0.25

    Find the differential equation of all the straight lines touching the circle x2  + y2  = r2 .

    Solutions

    1. Let y = mx + c be the equation of all the straight lines touching the circle.

      Given : The equation of the circle is x2  + y2  = r2 ---------->(1)

      The tangent to the circle is c2  = r2 (1+m2 )

      c = r √(1+m2 )

      we know that y = mx + c---------->(2)

      y = mx + r √(1+m2 ) ---------->(3)

      y - mx = r √(1+m2 )

      Differentiating wrt x we get dy/dx -m =0

      dy/dx = m

      Substituting this in equation (3)

      y - (dy/dx . x) = r √(1+(dy/dx)2 )

      Squaring on both sides, we get

      y2  - (dy/dx . x)2  = [ r √(1+(dy/dx)2 )]2

      [y - x(dy/dx)]2  = r2  (1+(dy/dx))2   is the required differential equation.

      Answer: The differential equation of all the straight lines touching the circle x2  + y2  = r2  is  [y - x(dy/dx)]2  = r2  (1+(dy/dx))2 ​​​​​​

     

     

  • Question 9/10
    1 / -0.25

    The solution of differential equation  x2 dy + y(x + y)dx = 0  when x = 1, y = 1 is:​

    Solutions

    x2 dy + (xy + y2 ) dx = 0
    ⇒ x2 dy = - (xy + y2 )dx
    ⇒  .......(i)
    Let y = vx
    Differentiating w.r.t x we get

    Substituting the value of y and dy/dx  in equation (1), we get:

    Integrating both sides, we get:


    ......(ii)
    Now, it is given that y = 1 at x = 1.

    Substituting D = 1/3  in equation (2), we get

  • Question 10/10
    1 / -0.25

    The solution of the differential equation is :

    Solutions

    None

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