Initial value problem (differential equations)?

Question

 Solve the initial value problem y'=2y+3 , y(0)=2

Thank you!  

 

 

 

Answer 1

Given  y'=2x+3 , y(0)=2

dy = (2x+3) dx

Apply integral each side.

 y = 2 × + 3x + c

y = x² +3x+c

Given y(0)=2

Now y(0) = 0² +3×0+c

y(0) = c⇒ c=2

y = x² +3x+2

The solution of  y'=2x+3 is  y = x² +3x+2.