If $\mathrm{f}^{\prime}(x)=x-\frac{5}{x^5}$ and $\mathrm{f}(1)=4$, then $\mathrm{f}(x)$ is

If $\mathrm{f}^{\prime}(x)=x-\frac{5}{x^5}$ and $\mathrm{f}(1)=4$, then $\mathrm{f}(x)$ is
  1. $\frac{x^2}{2}+\frac{9}{4} \frac{1}{x^4}+\frac{5}{4}$
  2. $\frac{x^2}{2}-\frac{5}{4} \frac{1}{x^4}+\frac{9}{4}$
  3. $\frac{x^2}{2}+\frac{5}{4} \frac{1}{x^4}+\frac{9}{4}$
  4. $\frac{x^2}{2}-\frac{9}{4} \frac{1}{x^4}+\frac{5}{4}$

Solution

Given $\mathrm{f}^{\prime}(x)=x-\frac{5}{x^5}$ $\therefore \quad$ Integrating both sides, we get $\begin{array}{ll} & \mathrm{f}(x)=\int\left(x-\frac{5}{x^5}\right) \mathrm{d} x \\ & \mathrm{f}(x)=\frac{x^2}{2}+\frac{5}{4} \times \frac{1}{x^4}+\mathrm{c} \\ \therefore \quad & \text { But } \mathrm{f}(1)=4 \\ \therefore \quad & \frac{1}{2}+\frac{5}{4}+\mathrm{c}=4 \\ \therefore \quad & \mathrm{c}=\frac{9}{4} \\ \therefore \quad & \mathrm{f}(x)=\frac{x^2}{2}+\frac{5}{4} \frac{1}{x^4}+\frac{9}{4} \end{array}$

Asked in: MHT CET 2023 (09 May Shift 2)

Practice more Differential Equations questions on Aicharya