Answer to Question #197055 in Differential Equations for Rabia

Question #197055

Identify the type of equations and Solve

a) š‘¦ā€² = š‘„š‘¦ 1+š‘„2

b) sin š‘¦ š‘‘š‘„ + cos š‘¦ š‘‘š‘¦ = 0

c) š‘‘š‘¦ = 2š‘” (š‘¦2 + 9 )š‘‘š‘”


1
Expert's answer
2021-05-28T06:10:05-0400

a)š‘¦ā€²=š‘„š‘¦1+š‘„2This linear ordinary Differential equation. Solving the given differential equation by variable separable form,dyy=š‘„1+š‘„2dx Integrate both side,āˆ«dyy=12āˆ«2š‘„1+š‘„2dxlny=12ln(1+x2)+cb)sinš‘¦š‘‘š‘„+cosš‘¦š‘‘š‘¦=0 This linear ordinary Differential equation. Solving the given differential equation by variable separable form,cosysinydy=āˆ’dx Integrate both side,āˆ«cosysinydy=āˆ’āˆ«dxlnsiny=āˆ’x+cc)š‘‘š‘¦=2š‘”(š‘¦2+9)š‘‘š‘” This linear ordinary Differential equation. Solving the given differential equation by variable separable form,š‘‘š‘¦=2š‘”(š‘¦2+9)š‘‘š‘”132+y2dy=2tdt1911+(y3)2dy=2tdt Integrate both side,19āˆ«11+(y3)2dy=2āˆ«tdt13tanāˆ’1(y3)=t2+ca)\newline š‘¦ā€² = \frac{š‘„š‘¦}{1+š‘„^2}\newline \text{This linear ordinary Differential equation.}\newline\text{ Solving the given differential equation by variable separable form,}\newline \frac{dy}{y}= \frac{š‘„}{1+š‘„^2}dx\newline \text{ Integrate both side,}\newline \int\frac{dy}{y}= \frac{1}{2}\int\frac{2š‘„}{1+š‘„^2}dx\newline lny= \frac{1}{2}ln(1+x^2)+c\newline b)\newline sin š‘¦ š‘‘š‘„ + cos š‘¦ š‘‘š‘¦ = 0\newline\text{ This linear ordinary Differential equation}.\newline\text{ Solving the given differential equation by variable separable form,}\newline \frac{cosy}{siny}dy= -dx\newline\text{ Integrate both side,}\newline \int\frac{cosy}{siny}dy= -\int dx\newline lnsiny=-x+c \newline c)\newline š‘‘š‘¦ = 2š‘” (š‘¦^2 + 9 )š‘‘š‘”\newline\text{ This linear ordinary Differential equation}.\newline\text{ Solving the given differential equation by variable separable form,}\newline š‘‘š‘¦ = 2š‘” (š‘¦^2 + 9 )š‘‘š‘”\newline \frac{1}{3^2+y^2}dy= 2tdt\newline \frac{1}{9} \frac{1}{1+(\frac{y}{3})^2}dy= 2tdt\newline\text{ Integrate both side,}\newline \frac{1}{9}\int \frac{1}{1+(\frac{y}{3})^2}dy= 2\int tdt\newline \frac{1}{3} tan^{-1}(\frac{y}{3})= t^2+c\newline


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