C' (X) = 2 + 0.15 X ; C(0) = 100 - Mathematics

प्रश्न

C' (x) = 2 + 0.15 x ; C(0) = 100

उत्तर

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

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Differential Equations

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Q 24 Q 23 Q 25

पाठ 22: Differential Equations - Exercise 22.05 [पृष्ठ ३४]

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आरडी शर्मा Mathematics [English] Class 12

पाठ 22 Differential Equations
Exercise 22.05 | Q 24 | पृष्ठ ३४

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