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

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

Question
Differential equations
asked 2021-01-08
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

Answers (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 \(\displaystyle{x}{\tan{{\left({y}\right)}}}-{2}{x}+{\ln{{y}}}={0}\)
0

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