Given that y=2 when x=-3,What is the value of y when x=5 for the differential equation y dy/dx =2x+3?
1 Answer
Jun 28, 2018
# y^2 = 2x^2 +6x + 4 #
# x=5 => y = +- 2sqrt(21) #
Explanation:
We have:
# ydy/dx=2x+3# with#y(-3)=2#
The DE is separable, so we simply separate the variables" to get:
# int \ y \ dy = int \ 2x+3 \ dx#
And Integrating we get:
# 1/2y^2 = x^2 +3x + C #
Using the initial condition,
# 1/2(2^2) = (-3)^2 +3(-3) + C #
# \ \ \ \ \ => 2 = 9 - 9 + C #
# \ \ \ \ \ => C=2 #
Thus, we gain the particular solution:
# 1/2y^2 = x^2 +3x + 2 #
# \ \ \ \ \ => y^2 = 2x^2 +6x + 4 #
And when
# y^2 = 2(5^2) +6(5) + 4 #
# \ \ \ \ =50 +30 + 4 #
# \ \ \ \ = 84 #
And so there are two solutions:
# y=+-sqrt(84) = +- 2sqrt(21) #