# solve the initial value problem: (tan(y)-2)dx+(xsec^2(y)+1/y)dy=0, y(0)=1

Question
Differential equations
solve the initial value problem: $$\displaystyle{\left({\tan{{\left({y}\right)}}}-{2}\right)}{\left.{d}{x}\right.}+{\left({x}{{\sec}^{{2}}{\left({y}\right)}}+\frac{{1}}{{y}}\right)}{\left.{d}{y}\right.}={0}$$, y(0)=1

2021-01-09
Given intial value problem is $$\displaystyle{\left({\tan{{\left({y}\right)}}}-{2}\right)}{\left.{d}{x}\right.}+{)}{x}{{\sec}^{{2}}{\left({y}\right)}}+\frac{{1}}{{y}}{)}{\left.{d}{y}\right.}={0},{y}{\left({0}\right)}={1}$$ Now,
$$\displaystyle{\left({\tan{{\left({y}\right)}}}-{2}\right)}{\left.{d}{x}\right.}+{\left({x}{{\sec}^{{2}}{\left({y}\right)}}+\frac{{1}}{{y}}\right)}{\left.{d}{y}\right.}={0}$$
$$\displaystyle\to{\left({\tan{{\left({y}\right)}}}{\left.{d}{x}\right.}+{x}{{\sec}^{{2}}{\left({y}\right)}}{\left.{d}{y}\right.}\right)}-{2}{\left.{d}{x}\right.}+{\left(\frac{{1}}{{y}}\right)}{\left.{d}{y}\right.}={0}$$
$$\displaystyle\to{d}{\left({x}{\tan{{\left({y}\right)}}}\right)}-{d}{\left({2}{x}\right)}+{d}{\left({\ln{{y}}}\right)}={0}$$
$$\displaystyle\to{d}{\left({x}{\tan{{\left({y}\right)}}}-{2}{x}+{\ln{{y}}}\right)}={0}$$
PSK->xtan(y)-2x+lny=c,
Now given that at x=0, y=1. So from the above solution we get, $$\displaystyle{x}{\tan{{\left({y}\right)}}}-{2}{x}+{\ln{{y}}}={c}$$
$$\displaystyle\to{0}\cdot{\tan{{\left({1}\right)}}}-{2}\cdot{0}+{\ln{{\left({1}\right)}}}={c}$$
$$\displaystyle\to{c}={\ln{{\left({1}\right)}}}={0}$$
Therefore, the general solution of the given intial value problem is $$\displaystyle{x}{\tan{{\left({y}\right)}}}-{2}{x}+{\ln{{y}}}={0}$$

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