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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

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

Given differential equation is:
$left parenthesis 1 plus straight x squared right parenthesis dy over dx plus straight y equals straight e to the power of tan to the power of negative 1 end exponent straight x end exponent rightwards double arrow dy over dx plus fraction numerator straight y over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction space equals space fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction$
This is a linear differential equation of the form
$dy over dx plus Py space equals space straight Q$
$where space straight P space equals space fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction space and space straight Q space equals fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction Therefore comma space straight I. straight F. space equals space straight e to the power of integral Pdx end exponent space equals space straight e to the power of tan to the power of negative 1 end exponent straight x end exponent Thus space the space solution space is comma straight y left parenthesis straight I. straight F right parenthesis space equals space integral straight Q thin space left parenthesis straight I. straight F right parenthesis space dx rightwards double arrow straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals integral fraction numerator straight e to the power of tan to the power of negative 1 end exponent straight x end exponent over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction cross times straight e to the power of tan to the power of negative 1 end exponent straight x end exponent dx Substitute space straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space equals space straight t semicolon$
$straight e to the power of tan to the power of negative 1 end exponent straight x end exponent space cross times fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis end fraction dx space equals space dt Thus comma space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals space integral tdt rightwards double arrow space space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals straight t squared over 2 plus straight C rightwards double arrow space straight y open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses space equals space open parentheses straight e to the power of tan to the power of negative 1 end exponent straight x end exponent close parentheses squared over 2 plus straight C$