Calculus Questions

1
Calculus Questions
Applied Mathematics

2
Calculus Questions
Task-1
a.i. Given : y = 4Asinx + 2Bcosx
Differentiating wrt x-
y’= 4Acosx – 2Bsinx
Again Differentiating wrt x-
y’’= - 4Asinx - 2Bcosx
y’’= -y or
y’’+ y = 0
Since we differentiate it twice, it is 2nd order differential equation
a.ii. Given : y = x + A
x
Differentiating wrt x-
y’= 1 - A
x2
y’= 1 - 1
x (y – x)
y’+ y
x - 1 = 1
y’+ y
x = 2
It is 1st Order Differential Equation
a.iii. Given : y = Ax2 – Bx (1)
Differentiating wrt x-
y’= 2Ax – B (2)
Again Differentiating wrt x-
y’’= 2A or A = y’’/2 (3)
Put (3) in (2)-

3
Calculus Questions
y’= y’’x – B or B = y’’x – y’ (4)
Put (3) and (4) in (1)-
y = y ' '
2 x2 – y’’x2 – y’x
y ' '
2 x2 + y’x + y = 0
It is a 2nd Order Differential Equation.
b.i. Given : dy
dx = y
x
dy
y = dx
x
Integrating both sides-
∫ dy
y =¿∫ dx
x ¿
ln(y) = ln(x) + ln(c) where ln(c) is the constant of integration
ln(y) = ln(xc) or
y = xc is the general solution of the given differential equation.
b.ii. Given : cosx
3+ y
dy
dx = sinx
Separating the variables-
dy
3+ y = tanx dx
Integrating both sides-
∫ dy
3+ y =∫tanx dx
ln(3 + y) = ln(secx) + ln(c) where ln(c) is the constant of integration
ln(3 + y) = ln(c*secx)
3 + y = c*secx