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Differential Equations

Short Answer Type

121. Find a one parameter family of solutions of each of the following differential equation:
(x y² + 2x) dx + (x² y + 2y) dy = 0

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122. Find a one parameter family of solutions of each of the following differential equation:
x y y' = 1 + x + y + x y

The given differential equation is

straight x space straight y space dy over dx space equals space 1 plus straight x plus straight y plus xy space space space space space space space space space or space space space space straight x space straight y dy over dx space equals space left parenthesis 1 plus straight x right parenthesis plus straight y left parenthesis 1 plus straight x right parenthesis

straight x space straight y dy over dx space equals space left parenthesis 1 plus straight x right parenthesis thin space left parenthesis 1 plus straight y right parenthesis

Separating the varibles, we get,

fraction numerator straight y over denominator 1 plus straight y end fraction dy space equals space fraction numerator 1 plus straight x over denominator straight x end fraction dx space space space space space space space or space space space space space space space space space integral fraction numerator straight y space dy over denominator 1 plus straight y end fraction space equals space integral fraction numerator 1 plus straight x over denominator straight x end fraction dx

therefore space space space space space integral open parentheses 1 minus fraction numerator 1 over denominator 1 plus straight y end fraction close parentheses space dy space equals space integral open parentheses 1 over straight x plus 1 close parentheses space dx therefore space space space straight y minus log space left parenthesis 1 plus straight y right parenthesis space equals space log space straight x space plus straight x space plus straight c comma space which space is space required space solution. space

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123.

Solve:
$dy over dx equals straight e to the power of straight x minus straight y end exponent plus straight x squared straight e to the power of straight y$

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124.

Solve:
$dy over dx space equals straight e to the power of straight x plus straight y end exponent plus straight x squared space straight e to the power of straight y.$

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125.

Solve:
$dy over dx space equals straight e to the power of straight x minus straight y end exponent plus straight x cubed straight e to the power of negative straight y end exponent$

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126.

Solve:
x (1 + y²) dx + y (1 + x²)dy = 0

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127.

Solve:
(1 + x) (1 + y²) dx + (1 + y) (1 + x²) dy = 0.

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128.

For the following differential equation, find the general solution:
$dy over dx equals fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction$