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Differential Equations Test - 3
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Differential Equations Test - 3
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  • Question 1/10
    1 / -0.25

    Write the order and degree of the given differential equation:

    Solutions

    Order = 3 and Degree = 1
    The order of a differential equation is determined by the highest-order derivative; the degree is determined by the highest power on a variable. The higher the order of the differential equation, the more arbitrary constants need to be added to the general solution.

  • Question 2/10
    1 / -0.25

    The order of a differential equation representing a family of curves is same as:​

    Solutions

    The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves.

  • Question 3/10
    1 / -0.25

    The solution of the equation x3 dx + (y + 1)2 dy = 0

    Solutions

    x3 dx + (y + 1)dy = 0
    =>d(x4 /4) + d((y + 1)3 /3) = 0
    =>d(x4 /4 + ((y + 1)3 /3)) = 0
    =>(x4 /4 + ((y + 1)3 /3)) = constt
    =>12 *(x4 /4 + ((y + 1)3 /3)) = 12 * constt
    =>3x4 + 4(y + 1)3 = constt

  • Question 4/10
    1 / -0.25

    Identify the type of differential equation

    Solutions

    (x −y)dy/dx = x + 2y
    ⇒dy/dx = (x + 2y)/(x −y)
    F (x,y) = (x + 2y)/(x −y)
    F(Ax, Ay) = (Ax + 2Ay)/(Ax −Ay) 
    = A(x + 2y)/A(x −y) 
    = (x + 2y)/(x −y) = F (x,y)
    Hence, the equation is homogenous.

  • Question 5/10
    1 / -0.25

    Identify the form of the given Differential Equation

  • Question 6/10
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    The solution of the initial value problem   is :

    Solutions

    xdy/dx = -coty
    dy/dx = - coty/x
    dy/coty = - dx/x
    tany dy = - dx/x  
    ∫(siny/cosy)dy = - ∫dx/x …………..(1)
    Put t = cosy
    dt = -sinydy
    Put the value of dt in eq(1)
    -∫dt/t = -∫dx/x
    log |t| = log |x| + log c
    log |cosy| = log(c.|x|)
    |cos y| = c.|x|....................(2)
    As y = π/4, x = (2)½
    cos π/4 = c*(2)½
    1/(2)½  = c*(2)½  
    c = ½
    Put the value of ‘c ’in eq(2)
    cos y = x/2   =>x = 2cosy

  • Question 7/10
    1 / -0.25

    The solution of the differential equation   is :

    Solutions

    dy/dx = eax cos by  
    ∫dy/cos by = ∫eax dx
    ∫sec by dy = ∫eax dx
    = (log| sec by + tan by|)/b = eax /a + c
    = a(log| sec by + tan by|) = beax + c

  • Question 8/10
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    The solution of the differential equation  is :

    Solutions

    dy/dx = x logx
    =>∫dy = ∫x logx dx
    y = logx . x2 /2 - ∫1/x . x2 /2 dx + c
    y = x2 /2 log x -½∫x dx + c
    y = x2 /2 log x - ½. x2 /2 + c
    y = x2 /2 log x - x2 /4 + c

  • Question 9/10
    1 / -0.25

    The solution of the differential equation   is :

    Solutions

     d2 y/dx2 = xex
    d2 y/dx2 =  ∫xex
    d/dx(dy/dx) = [xex ]
    Integrating, we get
    dy/dx = (x-1)ex + c1
    Again integrating, y = (x-2)ex + c1x + c2

  • Question 10/10
    1 / -0.25

    The solution of the initial value problem  edy/dx = x + 1, y(0) = 3 is :

    Solutions

     edy/dx = x + 1
    ⇒dy/dx = log(x+1)
    ⇒dy = log(x+1)dx
    Integrating both sides, we get
    ⇒∫dy =  ∫log(x+1)dx
    y = log(x + 1)  ∫1dx - ∫[d/dx{log(x + 1)} ∫1dx]dx
    y = xlog(x + 1) -  ∫1/(x+1)xdx
    y = xlog(x + 1) -  ∫(1 - 1/(x + 1)dx
    y = xlog(x + 1) -  ∫dx +  ∫1/(x+1)dx
    y = xlog(x + 1) - x + log|x + 1| + c
    y = (x + 1)log|x + 1| - x + c
    Here c is 3
    y = (x + 1)log|x + 1| - x + 3  

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