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Write the order and degree of the given differential equation:
Order = 3 and Degree = 1The 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.
The order of a differential equation representing a family of curves is same as:
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.
The solution of the equation x3 dx + (y + 1)2 dy = 0
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
Identify the type of differential equation
(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.
Identify the form of the given Differential Equation
The solution of the initial value problem is :
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
The solution of the differential equation is :
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
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
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
The solution of the initial value problem edy/dx = x + 1, y(0) = 3 is :
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 3y = (x + 1)log|x + 1| - x + 3
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