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Question Number 32308 by Ruchinna1 last updated on 22/Mar/18

Answered by Joel578 last updated on 23/Mar/18

$$\left(3\right)$$$$y^{⁗}=5x$$$$y^{‴}=\frac{5}{2}{x}^2+C$$$$y^{″}=\frac{5}{6}{x}^3+C_{1}x+C_2$$$$y^{\prime }=\frac{5}{24}{x}^4+\frac{C_1}{2}{x}^2+C_{2}x+C_3$$$$y=\frac{1}{24}{x}^5+\frac{1}{6}{C}_1{x}^3+\frac{1}{2}{C}_2{x}^2+C_{3}x+C_4$$

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