Question

If the general solution of the differential equation $\left(\text{y}\right)^{'}=\frac{\text{y}}{\text{x}}+\phi\left(\frac{\text{x}}{\text{y}}\right)$ , for some function $\phi$ , is given by $yln\left|\right.\text{c}\text{x}\left|\right.=\text{x}$ , where $\text{c}$ is an arbitrary constant, then $\phi\left(2\right)$ is equal to $\left(\right)\text{here} , \, \left(\text{y}\right)^{'}=\frac{d y}{d x}\left(\text{y}\right)^{'}=\frac{d y}{d x}$

• A. $-4$
• B. $- \frac{1}{4}$
• C. $\frac{1}{4}$
• D. $4$

Answer 1

Answer: (B)

$- \frac{1}{4}$

Answer 2

$\text{given} \text{:} \left(\text{y}\right)^{'} = \frac{\text{y}}{\text{x}} + \phi \left(\frac{\text{x}}{\text{y}}\right)$ As $\text{y ln} \left(\text{cx}\right) = \text{x} \, ? \, \left(\text{y}\right)^{'} \text{ln} \left(\text{cx}\right) + \text{y} \frac{1}{\text{cx}} \text{c} = 1$ $? \, \left(\text{y}\right)^{'} \left(\frac{\text{x}}{\text{y}}\right) + \frac{\text{y}}{\text{x}} = 1$ $? \, \, \, \frac{1 - \left(\frac{\text{y}}{\text{x}}\right)}{\left(\frac{\text{x}}{\text{y}}\right)} = \left(\frac{\text{y}}{\text{x}}\right) + \phi \left(\frac{\text{x}}{\text{y}}\right)$ Put $\frac{x}{y} = 2 \, ? \, \frac{1 - \left(\frac{1}{2}\right)}{\left(\frac{2}{1}\right)} = \left(\frac{1}{2}\right) + \phi \left(\frac{2}{1}\right)$ $? \, \, \, \phi \left(2\right) = - \frac{1}{4}$