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Differential Equations

Multiple Choice Questions

301. The general solution of the differential equation e^x dy + (ye^x + 2x) dx = 0 is

x e^y + x² = C
x e^y + y² = C
y e^x + x² = C
y e^x + x² = C

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Short Answer Type

302.

Find the particular solution of the differential equation
$dy over dx equals negative fraction numerator x plus y space c o s space x over denominator 1 plus s i n space x space end fraction space g i v e n space t h a t space y equals 1 space w h e n space x equals 0$

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303.

Find the particular solution of the differential equation

2ye^x/y dx+ (Y-2xe^x/y) dy =0

Given that x=0 when y=1.

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304.

Find the differential equation of the family of lines passing through the origin.

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305. $If space straight y space equals space straight e to the power of ax. cosbx comma space then space prove space that space space space fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 straight a dy over dx plus left parenthesis straight a squared plus straight b squared right parenthesis straight y space equals space 0$

straight y equals straight e to the power of ax. cos space bx dy over dx equals ae to the power of ax. cosbx minus be to the power of ax. sinbx space space... left parenthesis straight i right parenthesis dy over dx equals ay minus be to the power of ax. space sinbx fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight a dy over dx minus straight b left parenthesis ae to the power of ax. sinbx plus be to the power of ax. cosbx right parenthesis fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight a dy over dx minus abe to the power of ax. sinbx minus straight b squared straight e to the power of ax. cos space bx fraction numerator straight d squared straight y over denominator dx squared end fraction equals straight a dy over dx minus straight a open parentheses ay minus dy over dx close parentheses minus straight b squared straight y space left square bracket Substituting space be to the power of ax space sinbx space from space left parenthesis straight i right parenthesis right square bracket

fraction numerator straight d squared straight y over denominator dx squared end fraction space equals straight a dy over dx minus straight a squared straight y plus straight a dy over dx minus straight b squared straight y therefore space space space fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 straight a dy over dx plus left parenthesis straight a squared plus straight b squared right parenthesis straight y space equals space 0

Hence Proved

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306.

If space straight x to the power of straight x plus straight x to the power of straight y plus straight y to the power of straight x space equals space straight a to the power of straight b comma space space then space find space dy over dx.

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307.

If $straight x equals straight a space sin space 2 straight t left parenthesis 1 plus cos space 2 straight t right parenthesis space and space straight y space equals straight b space cos space 2 straight t left parenthesis 1 minus cos space 2 straight t right parenthesis comma space then space find dy over dx space at space straight t space equals straight pi over 4.$

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Long Answer Type

308.

Find the particulars solution of the differential equation $straight x squared dy space equals space left parenthesis 2 xy plus straight y squared right parenthesis dx comma space$ given that y =1 when x = 1.

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309.

Find the particulars solution of the differential equation $left parenthesis 1 plus straight x squared right parenthesis dy over dx equals open parentheses straight e to the power of straight m space tan to the power of negative 1 straight x end exponent end exponent minus straight y close parentheses comma space given space that space straight y space equals space 1 space when space straight x space equals space 0.$

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Short Answer Type

310.

Solve the differential equation $left parenthesis 1 plus straight x squared right parenthesis dy over dx plus straight y equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent$

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