# 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) #