हिंदी
तमिलनाडु बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी वाणिज्य कक्षा १२

Solve the following differential equation: Suppose that the quantity demanded Qd = pdpdtdpdt13-6p+2dpdt+d2pdt2 and quantity supplied Qs = p-3+2p where p is the price. Find the - Business Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Solve the following differential equation:

Suppose that the quantity demanded Qd = `13 - 6"p" + 2 "dp"/"dt" + ("d"^2"p")/("dt"^2)` and quantity supplied Qs = `- 3 + 2"p"` where p is the price. Find the equilibrium price for market clearance

योग

उत्तर

For market clearance, the required condition is Qd = Qs

`13 - 6"p" + 2 "dp"/"dt" + ("d"^2"p")/("dt"^2) = - 3 + 2"p"`

`("d"^2"p")/("dt"^2) + 2 "dp"/"dt" - 6"p" + 13 + 3 = 0`

`("d"^2"p")/("dt"^2) + 2 "dp"/"dt" - 8"p" = - 16`

`("D"^2 + 2"D" - 8)"p" = - 16`

The auxiliary equation is

m2 + 2m – 8 = 0

(m + 4)(m – 2) = 0

m = – 4, 2

Roots are real and different

C.F = Aem1t + Bem2t

C.F = `"Ae"^(- 4"t") + "BE"^(2"t")`

P.I = `1/(("D"^2 + 2"D" - 8)) (- 16)`

= `1/(("D"^2 + 2"D" - 8)) (- 16 "e"^(0_x))`

= `1/((0 + 2(0) - 8)) (- 16 "e"^(0_x))`

= `(-16)/(- 8)`

P.I = 2

The general solution is y = C.F + P.I

P = `"Ae"^(-4"t") + "Be"^(2"t") + 2`

shaalaa.com
Second Order First Degree Differential Equations with Constant Coefficients
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Differential Equations - Exercise 4.5 [पृष्ठ ९९]

APPEARS IN

सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
अध्याय 4 Differential Equations
Exercise 4.5 | Q 13 | पृष्ठ ९९
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×