Question

If dy/dx= x-x²/2x⁴ and y=2 when x=1 to express y in terms of x.

Answer 1

$$\begin{gathered} \int \frac{x-x^{2}}{2 x^{4}} \mathrm{~d} x \\ =\int\left(\frac{1}{2 x^{3}}-\frac{1}{2 x^{2}}\right) \mathrm{d} x \\ =\frac{1}{2} \int \frac{1}{x^{3}} \mathrm{~d} x-\frac{1}{2} \int \frac{1}{x^{2}} \mathrm{~d} x \end{gathered} \\[4mm] \text { Now solving: }\\[2mm] \int \frac{1}{x^{3}} d x\\[2mm] \text{Apply power rule:}\\[2mm] \begin{aligned} \int x^{\mathrm{n}} \mathrm{d} x=& \frac{x^{\mathrm{n}+1}}{\mathrm{n}+1} \text { with } \mathrm{n}=-3: \\ &=-\frac{1}{2 x^{2}} \end{aligned} \\[2mm] \text{ Now solving:}\\[2mm] \int \frac{1}{x^{2}} \mathrm{~d} x\\[2mm] \text{ Apply power rule with } n =-2 :\\[2mm] =-\frac{1}{x}\\[3mm] \text{ Plug in solved integrals:}\\[3mm] \begin{gathered} \frac{1}{2} \int \frac{1}{x^{3}} \mathrm{~d} x-\frac{1}{2} \int \frac{1}{x^{2}} \mathrm{~d} x \\ =\frac{1}{2 x}-\frac{1}{4 x^{2}} \end{gathered} \\[2mm] \text{ The problem is solved:}\\[2mm] \begin{aligned} & \int \frac{x-x^{2}}{2 x^{4}} \mathrm{~d} x \\ =& \frac{1}{2 x}-\frac{1}{4 x^{2}}+C \end{aligned} \\[3mm] \text{ Simplifying the above, we have:}\\[2mm] y=\frac{2 x-1}{4 x^{2}}+C$$

At y=2 when x=1, we have:

$$2=\frac{2 (1)-1}{4 (1)^{2}}+C\\[1.5mm] 2 = \frac{2 -1}{4 }+C\\[1.5mm] 2 = \frac{1}{4 }+C\\[1.5mm] 2 -\frac{1}{4 } = C\\[2mm] \implies C = 1.75\\[4mm] \therefore y=\frac{2x-1}{4x^{2}}+1.75$$