Solve the following differential equation: from Mathematics Di
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    Differential Equations

    Multiple Choice QuestionsShort Answer Type

    181.

    Solve:
    cos space left parenthesis straight x plus straight y right parenthesis space dy over dx space equals space 1.

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    Answer

    Multiple Choice QuestionsLong Answer Type

    182.

    Solve:
    dy over dx space equals space tan space left parenthesis straight x plus straight y right parenthesis


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    Answer
    183. Solve the following initial value problem:
    (x + y + 1)2 dy = dx, y ( –1) = 0
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    Answer
    184.

    Solve the following initial value problem
    cos (x + y) dy = dx, y (0) = 0

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    Answer
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    185. For the following differential equation, given below, find  particular solution satisfying the given condition:
    open parentheses straight x cubed plus straight x squared plus straight x plus 1 close parentheses dy over dx space equals space 2 straight x squared plus straight x semicolon space space straight y space equals space 1 space space when space straight x space equals space 0
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    Answer
    186. For the following differential equation, given below, find  particular solution satisfying the given condition:
    straight x open parentheses straight x squared minus 1 close parentheses space dy over dx space equals space 1 space semicolon space space straight y space equals space 0 space space when space straight x space equals space 2
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    Answer

    Multiple Choice QuestionsShort Answer Type

    187. For the following differential equation, given below, find  particular solution satisfying the given condition:
    cos space open parentheses dy over dx close parentheses space space equals straight a space space space left parenthesis straight a space element of space straight R right parenthesis semicolon space space space straight y space equals space 1 space space when space straight x space equals space 0
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    Answer
    188. For the following differential equation, given below, find  particular solution satisfying the given condition:
    dy over dx equals straight y space tanx space semicolon space space straight y space equals space 1 space space when space straight x space equals space 0
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    Answer
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    Multiple Choice QuestionsLong Answer Type

    189.

    Solve:   straight x squared dy over dx space equals straight x squared plus 5 xy plus 4 straight y squared.

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    Answer
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    190.

    Solve the following differential equation:
    straight x squared dy over dx space equals space 2 xy plus straight y squared

    The given differential equation is
                                     straight x squared dy over dx space equals space 2 xy plus straight y squared
    or         dy over dx space equals space fraction numerator 2 xy plus straight y squared over denominator straight x squared end fraction                                ...(1)
    Put         straight y space equals space straight v space straight x space space space space so space that space space space dy over dx equals straight v plus dv over dx

    therefore space space space from space left parenthesis 1 right parenthesis comma space space straight v plus straight x dv over dx space equals space fraction numerator 2 straight x. space vx plus straight v squared straight x squared over denominator straight x squared end fraction
    or    straight v plus straight x space dv over dx space equals space 2 space straight v space plus space straight v squared
    or     straight x dv over dx space equals straight v squared plus straight v                     
    Separating the variables, we get,
                            fraction numerator 1 over denominator straight v squared plus straight v end fraction dv space equals space 1 over straight x dx

    Integrating,   integral fraction numerator 1 over denominator straight v squared plus straight v end fraction dv space equals space integral 1 over straight x dx
    therefore space space space integral fraction numerator 1 over denominator straight v left parenthesis straight v plus 1 right parenthesis end fraction dv space equals space integral 1 over straight x dx

    space therefore space space space space integral open square brackets fraction numerator 1 over denominator straight v left parenthesis 0 plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis negative 1 right parenthesis thin space left parenthesis straight v plus 1 right parenthesis end fraction close square brackets space dv space equals space integral space 1 over straight x dx therefore space space space space integral open parentheses 1 over straight v minus fraction numerator 1 over denominator straight v plus 1 end fraction close parentheses space dv space equals space integral 1 over straight x dx therefore space space space log space open vertical bar straight v close vertical bar space minus space log space open vertical bar straight v plus 1 close vertical bar space equals space log space open vertical bar straight x close vertical bar space plus space log space straight c subscript 1 therefore space space log space open vertical bar fraction numerator straight v over denominator straight v plus 1 end fraction close vertical bar space equals space log space left parenthesis straight c subscript 1 space open vertical bar straight x close vertical bar right parenthesis therefore space space open vertical bar fraction numerator straight v over denominator straight v plus 1 end fraction close vertical bar space equals space straight c subscript 1 open vertical bar straight x close vertical bar therefore space space space space space fraction numerator begin display style straight y over straight x end style over denominator begin display style straight y over straight x plus 1 end style end fraction equals space space cx therefore space space space space fraction numerator straight y over denominator straight x plus straight y end fraction space equals space straight c space straight x comma space which space is space required space solution. space
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