Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Differential Equations

Short Answer Type

191.

Solve the differential equation:
$straight x squared dy over dx space equals straight y left parenthesis straight x plus straight y right parenthesis.$

76 Views

Long Answer Type

192.

Solve the differential equation:
$left parenthesis straight y plus straight x right parenthesis space dy over dx space equals space straight y minus straight x.$

71 Views

193.
Solve the following differential equation:
(y² – x²) dy = 3 x y dx

The given differential equation is

open parentheses straight y squared minus straight x squared close parentheses space dy space equals space 3 xy space dx

dy over dx space equals space fraction numerator 3 space straight x space straight y over denominator straight y squared minus straight x squared end fraction

...(1)
Put y = v x so that

dy over dx space equals straight v plus straight x dv over dx.

therefore space space space space space space space left parenthesis 1 right parenthesis space becomes comma space space straight v plus straight x dv over dx space equals space fraction numerator 3 straight x. space straight v space straight x over denominator straight v squared straight x squared minus straight x squared end fraction

straight v plus straight x dv over dx space equals space fraction numerator 3 straight v over denominator straight v squared minus 1 end fraction

therefore space space space space space space straight x space dv over dx space equals space fraction numerator 3 straight v over denominator straight v squared minus 1 end fraction minus straight v space space space space or space space space space straight x space dv over dx space equals space fraction numerator 4 straight v space minus straight v cubed over denominator straight v squared minus 1 end fraction

Separating the variables, we get,

fraction numerator straight v squared minus 1 over denominator 4 space straight v minus straight v cubed end fraction dv space equals space 1 over straight x dx space space space or space space space fraction numerator straight v squared minus 1 over denominator straight v left parenthesis 4 minus straight v squared right parenthesis end fraction space dv space equals space 1 over straight x dx

Integrating,

integral fraction numerator straight v squared minus 1 over denominator straight v space left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 2 plus straight v right parenthesis end fraction dv space equals space integral 1 over straight x dx

therefore space space space space integral open square brackets fraction numerator 0 minus 1 over denominator straight v space left parenthesis 2 minus 0 right parenthesis thin space left parenthesis 2 plus 0 right parenthesis end fraction minus fraction numerator 4 minus 1 over denominator left parenthesis 2 right parenthesis thin space left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 2 plus 2 right parenthesis end fraction plus fraction numerator 4 minus 1 over denominator left parenthesis negative 2 right parenthesis thin space left parenthesis 2 plus 2 right parenthesis thin space left parenthesis 2 plus straight v right parenthesis end fraction close square brackets dv space equals space integral 1 over straight x dx

therefore space space integral open square brackets negative fraction numerator 1 over denominator 4 straight v end fraction plus fraction numerator 3 over denominator 8 left parenthesis 2 minus straight v right parenthesis end fraction minus fraction numerator 3 over denominator 8 left parenthesis 2 plus straight v right parenthesis end fraction close square brackets space dv space equals space integral 1 over straight x dx therefore space space space space minus space 1 fourth open vertical bar straight v close vertical bar plus space 3 over 8 fraction numerator log space open vertical bar 2 minus straight v close vertical bar over denominator negative 1 end fraction space minus space 3 over 8 space log space open vertical bar 2 plus straight v close vertical bar space equals space log space straight x space plus straight c therefore space space space space space minus 1 fourth space log space open vertical bar straight y over straight x close vertical bar space minus space 3 over 8 space log space open vertical bar 2 minus straight y over straight x close vertical bar space minus space 3 over 8 log space open vertical bar 2 plus straight y over straight x close vertical bar space equals space logx plus straight c

which is required solution.

74 Views

194. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(x - y) y' = x + 2 y

78 Views

195. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(x² + y²) y' = 8 x² - 3 x y + 2 y²

74 Views

196. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(3 x y + y²) dx = (x² + x y) dy

71 Views

197. Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
2 x y dx + (x² + 2 y²) dy = 0

74 Views