Question
Solve the following differential equation
$\text{C}(\text{x})=2+0.15\text{x},\text{C}(0)=100$

Answer

$\text{C}(\text{x})=2+0.15\text{x},\text{C}(0)=100$$\text{C}'(\text{x})\text{dx}=(2+0.15\text{x})\text{dx}$
$\int\text{C}'(\text{x})\text{dx}=\int2\text{dx}+0.15\int\text{x dx}$
$\text{C}(\text{x})=2\text{x}+0.15\frac{\text{x}^2}{2}+\text{C}\ ...(1)$
Put x = 0, c(x) = 100
100 = 2(0) + 0 + c
100 = c
Put c = 100 in equation 1
$\text{c}(\text{x})=2\text{x}+(0.15)\frac{\text{x}^2}{2}+100$

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