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    Differential Equations
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    Solve the following differential equation:
    left parenthesis 1 minus straight x squared right parenthesis space dy space plus space straight x space straight y space dx space equals space straight x space straight y squared space dx.
                
    The given differential equation is
                         left parenthesis 1 minus straight x squared right parenthesis space dy space plus space straight x space straight y space dx space equals space straight x space straight y squared space dx
    or                  left parenthesis 1 minus straight x squared right parenthesis space dy space equals space left parenthesis xy squared minus straight x space straight y right parenthesis space dx space space space space space space space or space space space left parenthesis 1 minus straight x squared right parenthesis space dy space equals space straight x space left parenthesis straight y squared minus straight y right parenthesis space dx
    or                 fraction numerator 1 over denominator straight y squared minus straight y end fraction dy space equals space fraction numerator straight x over denominator 1 minus straight x squared end fraction dx
    therefore space space space space space integral fraction numerator 1 over denominator straight y squared minus straight y end fraction dy space equals space integral fraction numerator straight x over denominator 1 minus straight x squared end fraction dx therefore space space space space integral fraction numerator 1 over denominator straight y left parenthesis straight y minus 1 right parenthesis end fraction dy space equals space minus integral fraction numerator negative 2 straight x over denominator 1 minus straight x squared end fraction dx therefore space space integral open square brackets fraction numerator 1 over denominator straight y left parenthesis 0 minus 1 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis 1 right parenthesis space left parenthesis straight y minus 1 right parenthesis end fraction close square brackets space dy space equals space minus integral fraction numerator negative 2 straight x over denominator 1 minus straight x squared end fraction dx therefore space integral open parentheses negative 1 over straight y plus fraction numerator 1 over denominator straight y minus 1 end fraction close parentheses space dy space equals negative integral fraction numerator negative 2 straight x over denominator 1 minus straight x squared end fraction dx therefore space space space space minus log space open vertical bar straight y close vertical bar plus space log space open vertical bar straight y minus 1 close vertical bar space equals space minus space log space open vertical bar 1 minus straight x squared close vertical bar plus space log space straight c subscript 1 therefore space log space open vertical bar fraction numerator straight y minus 1 over denominator straight y end fraction close vertical bar plus space log space open vertical bar 1 minus straight x squared close vertical bar space equals space log space straight c subscript 1 therefore space log space open vertical bar open parentheses fraction numerator straight y minus 1 over denominator straight y end fraction close parentheses space left parenthesis 1 minus straight x squared right parenthesis close vertical bar space equals space log space straight c subscript 1 space space space space space space space space space therefore space space space space open vertical bar fraction numerator left parenthesis straight y minus 1 right parenthesis thin space left parenthesis 1 minus straight x squared right parenthesis over denominator straight y end fraction close vertical bar space equals space straight c subscript 1 therefore space space space space left parenthesis straight y minus 1 right parenthesis thin space left parenthesis 1 minus straight x squared right parenthesis space equals space straight c space straight y
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    Answer
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    Solve the following differential equation:
    x cos y dy = (x ex log x + ex) dx.

    The given differential equation is
                     straight x space cos space straight y space dy space equals space left parenthesis straight x space straight e to the power of straight x space log space straight x space plus straight e to the power of straight x right parenthesis space dx
    or              cos space straight y space dy space equals space open parentheses straight e to the power of straight x logx plus straight e to the power of straight x over straight x close parentheses dx
    Integrating,     integral space cos space straight y space dy space equals space integral space straight e to the power of straight x open square brackets log space straight x space plus 1 over straight x close square brackets space dx
    therefore               space sin space straight y space equals space straight e to the power of straight x logx space plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open curly brackets because space space space integral straight e to the power of straight x left square bracket straight f left parenthesis straight x right parenthesis plus straight f apostrophe left parenthesis straight x right parenthesis dx space equals space straight e to the power of straight x space straight f left parenthesis straight x right parenthesis close curly brackets
    which is the required solution. 
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    Solve the following differential equation:
                left parenthesis straight y plus straight x space straight y right parenthesis space dx space plus space left parenthesis straight x minus straight x space straight y squared right parenthesis space dy space equals space 0

    The given differential equation is
                      left parenthesis straight y plus straight x space straight y right parenthesis space dx space plus space left parenthesis straight x minus straight x space straight y squared right parenthesis space dy space equals space 0
    or space straight y left parenthesis 1 plus straight x right parenthesis space dx space plus space straight x left parenthesis 1 minus straight y squared right parenthesis space dy space equals space 0 or space space straight x left parenthesis 1 minus straight y squared right parenthesis space dy space equals space minus straight y left parenthesis 1 plus straight x right parenthesis space dx or space fraction numerator 1 minus straight y squared over denominator straight y end fraction dy space equals space minus fraction numerator 1 plus straight x over denominator straight x end fraction dx
    Integrating,       integral fraction numerator 1 minus straight y squared over denominator straight y end fraction dy space equals space minus fraction numerator 1 plus straight x over denominator straight x end fraction dx
    therefore space space space integral open parentheses 1 over straight y minus straight y close parentheses space dy space equals space minus integral open parentheses 1 over straight x plus 1 close parentheses space dx therefore space space log space open vertical bar straight y close vertical bar minus straight y squared over 2 space equals space minus log space straight x space minus space straight x space plus straight c comma space which space is space the space required space solution. space

    86 Views
    Answer

    Solve the differential equation
    left parenthesis tan squared straight x plus 2 space tanx space plus 5 right parenthesis space dy over dx space equals space 2 left parenthesis 1 plus tanx right parenthesis space sec squared straight x.

    The given differential equation is
                    open parentheses tan squared straight x plus 2 space tan space straight x plus 5 close parentheses space dy over dx space equals space 2 left parenthesis 1 minus tanx right parenthesis space sec squared straight x
    or        dy over dx equals space fraction numerator 2 space sec squared straight x plus 2 space tanx space sec squared straight x over denominator tan squared straight x plus 2 space tanx plus 5 end fraction
    Separating the variables,  we get,         
                          dy space equals space fraction numerator 2 space tanx space sec squared straight x space plus space 2 space sec squared straight x over denominator tan squared straight x plus 2 space tan space straight x plus 5 end fraction dx
    Integrating ,   integral 1 space dy space equals space integral fraction numerator 2 space tanx space sec squared straight x space plus space 2 sec squared straight x over denominator tan squared straight x plus 2 space tan space straight x plus 5 end fraction dx
    therefore space space space space space space space space space space space straight y equals log space open vertical bar tan squared straight x plus 2 tanx plus 5 close vertical bar space plus straight c space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space space integral fraction numerator straight f apostrophe left parenthesis straight x right parenthesis over denominator straight f left parenthesis straight x right parenthesis end fraction dx space equals space log space open vertical bar straight f left parenthesis straight x right parenthesis close vertical bar close square brackets
    which is the required solution. 
     
    85 Views
    Answer
    Solve the following differential equation:
    dy over dx space equals space log space left parenthesis straight x plus 1 right parenthesis
    The given differential equation is
                          dy over dx space equals space log space left parenthesis straight x plus 1 right parenthesis
    Separating the variables, we get,
                            dy space equals space log space left parenthesis straight x plus 1 right parenthesis. space dx
    Integrating,       integral dy space equals space integral log space left parenthesis straight x plus 1 right parenthesis. space 1 space dx
    therefore space space space space space space space space space space space straight y space equals space log space left parenthesis straight x plus 1 right parenthesis space. straight x space space minus space integral fraction numerator 1 over denominator straight x plus 1 end fraction straight x space dx

    or            straight y equals straight x space log left parenthesis straight x plus 1 right parenthesis minus space integral fraction numerator left parenthesis straight x plus 1 right parenthesis space minus space 1 over denominator straight x plus 1 end fraction dx

    or            straight y space equals straight x space log space left parenthesis straight x plus 1 right parenthesis space minus space integral open parentheses 1 minus fraction numerator 1 over denominator straight x plus 1 end fraction close parentheses dx

    or           straight y equals straight x space log space left parenthesis straight x plus 1 right parenthesis space minus space open square brackets straight x minus log left parenthesis straight x plus 1 right parenthesis close square brackets plus straight c

    or          straight y equals space straight x space log space left parenthesis straight x plus 1 right parenthesis space minus space straight x space plus space log space left parenthesis straight x plus 1 right parenthesis space plus straight c
    or           straight y equals left parenthesis straight x plus 1 right parenthesis space log space left parenthesis straight x plus 1 right parenthesis space minus straight x space plus straight c
    which is required solution. 
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    Answer
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