Question

C' (x) = 2 + 0.15 x ; C(0) = 100

Answer 1

We have, 
\[C' \left( x \right) = 2 + 0 . 15x\]
\[ \Rightarrow \frac{dC}{dx} = 2 + 0 . 15x\]
\[ \Rightarrow dC = \left( 2 + 0 . 15x \right)dx\]
Integrating both sides, we get
\[\int dC = \int\left( 2 + 0 . 15x \right) dx\]
\[ \Rightarrow C = 2x + \frac{0 . 15}{2} x^2 + D . . . . . \left( 1 \right)\]
\[\text{ It is given that C }\left( 0 \right) = 100 . \]
\[ \therefore 100 = 2\left( 0 \right) + \frac{0 . 15}{2}\left( 0 \right) + D\]
\[ \Rightarrow D = 100\]
\[\text{ Substituting the value of D in } \left( 1 \right), \text{ we get }\]
\[C = 2x + \frac{0 . 15}{2} x^2 + 100\]
\[\text{ Hence, }C = 2x + \frac{0 . 15}{2} x^2 + 100 \text{ is the solution to the given differential equation .}\]