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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

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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

The given differential equation is
x(1 + y²) dx + y(1 + x²)dy = 0
$therefore space space space space straight y left parenthesis 1 plus straight x squared right parenthesis space dy equals space minus straight x left parenthesis 1 plus straight y squared right parenthesis space dx therefore space space space space space fraction numerator straight y over denominator 1 plus straight x squared end fraction dy space equals space minus fraction numerator straight x over denominator 1 plus straight x squared end fraction dx$
Integrating, $integral fraction numerator straight y over denominator 1 plus straight y squared end fraction dy space equals space minus integral fraction numerator straight x over denominator 1 plus straight x squared end fraction dx$
$therefore space space space space integral fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy plus integral fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction dx space equals space 0$
$therefore space space space log space left parenthesis 1 plus straight y squared right parenthesis plus space log space left parenthesis 1 plus straight x squared right parenthesis space equals space log space straight c therefore space space log space left parenthesis 1 plus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis space equals space log space straight c therefore space space left parenthesis 1 plus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis space equals straight c$
which is required solution.

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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:
$dy over dx equals fraction numerator 1 minus cosx over denominator 1 plus cosx end fraction$