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Differential Equations

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

The given differential equation is
(x + y + 1)²dy = dx
or

dy over dx space equals space fraction numerator 1 over denominator left parenthesis straight x plus straight y plus 1 right parenthesis squared end fraction

Put

straight x plus straight y plus 1 space equals space straight t comma space space space space therefore space space space space 1 plus dy over dx space equals dt over dx space space space or space space space dy over dx equals space dt over dx minus 1

therefore space space space space given space equation space becomes space dt over dx minus 1 space equals 1 over straight t squared

therefore space space space space space space space space space space space dt over dx equals 1 plus 1 over straight t squared

dt over dx space equals fraction numerator straight t squared plus 1 over denominator straight t squared end fraction

Separating the variables and integrating, we get,

integral fraction numerator straight t squared over denominator straight t squared plus 1 end fraction dt space equals space integral dx space space space space or space space space integral open parentheses 1 minus fraction numerator 1 over denominator 1 plus straight t squared end fraction close parentheses space dt space equals space integral 1. space dx

therefore space space space space space space straight t minus tan to the power of negative 1 end exponent straight t space equals space straight x plus straight c space space space space or space space space left parenthesis straight x plus straight y plus 1 right parenthesis space minus space tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space straight x plus straight c therefore space space space space straight y plus 1 minus tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space straight c

Now

straight y left parenthesis negative 1 right parenthesis space equals space 0 space space space space space space space space rightwards double arrow space space space space straight y space equals space 0 space space space when space straight x space equals space minus 1

therefore space space space space 0 plus 1 space minus space tan to the power of negative 1 end exponent left parenthesis negative 1 plus 0 plus 1 right parenthesis space equals space straight c space space space space space space rightwards double arrow space space space space 1 minus tan to the power of negative 1 end exponent 0 space equals space straight c therefore space space space space space space space space space space space space space space 1 minus 0 space equals space straight c space space space space space space space space rightwards double arrow space space space space straight c space equals space 1 therefore space space space space space from space left parenthesis 1 right parenthesis comma space space straight y plus 1 space minus space tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space 1 therefore space space tan to the power of negative 1 end exponent left parenthesis straight x plus straight y plus 1 right parenthesis space equals space straight y space space space space space or space space space straight x plus straight y plus 1 space equals space tan space straight y

which is required solution.

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184.

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

74 Views

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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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

80 Views

Short 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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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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Long Answer Type

189.

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

80 Views

190.

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

74 Views

Previous Year Papers

Test Series

Test Yourself

Short Answer Type

Long Answer Type

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

Short Answer Type

Long Answer Type

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