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Differential Equations

Long Answer Type

281. Find the general solution of the following differential equation:

dy over dx plus fraction numerator 4 straight x over denominator straight x squared plus 1 end fraction straight y space equals negative fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis cubed end fraction

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Short Answer Type

282.
Solve: $open parentheses straight x plus straight y close parentheses space dy over dx space equals space 1.$

The given differential equation is

open parentheses straight x plus straight y close parentheses dy over dx equals 1 space space space space space space space space space space or space space space straight x plus straight y space equals dy over dx space space space or space space space dx over dy minus straight x space equals space straight y

Comparing it with

dx over dy plus Py space equals straight Q comma space space we space get comma space straight P space equals space minus 1 comma space space straight Q space equals space straight y

integral straight P space dy space equals space minus integral 1 space dy space equals space minus straight y straight e to the power of integral straight P space dy end exponent space equals space straight e to the power of integral space 1 space dy end exponent space equals space straight e to the power of negative straight y end exponent

therefore

solution of given equation is

straight x. space straight e to the power of integral straight P space dy end exponent space equals space integral straight Q. space straight e to the power of integral straight P space dy end exponent space dy space plus straight c

straight x. space straight e to the power of negative straight y end exponent space equals space integral straight y. space straight e to the power of negative straight y end exponent dy plus straight c

or space space space space straight x space straight e to the power of negative straight y end exponent space equals space straight y fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction space minus space integral 1. space fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction dy plus straight c or space space space straight x space straight e to the power of negative straight y end exponent space equals space minus ye to the power of negative straight y end exponent plus integral straight e to the power of negative straight y end exponent space dy space plus space straight c or space space space xe to the power of negative straight y end exponent space equals space minus ye to the power of negative straight y end exponent plus fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction plus straight c or space space space xe to the power of negative straight y end exponent space equals space minus ye to the power of negative straight y end exponent plus fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction plus straight c or space space space space xe to the power of negative straight y end exponent space equals space minus ye to the power of negative straight y end exponent minus straight e to the power of negative straight y end exponent plus straight c or space space space space space space space space space straight x space equals space minus straight y space minus space 1 space plus space straight c space straight e to the power of straight y or space space space space straight x plus straight y plus 1 space equals space ce to the power of straight y

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Long Answer Type

283.

Solve:
$open parentheses straight x plus 2 straight y cubed close parentheses space straight y apostrophe space equals space straight y comma space space straight y greater than 0$

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Short Answer Type

284.

Solve:
$straight y space dx space plus space left parenthesis straight x minus straight y right parenthesis squared space dy space equals space 0$

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Long Answer Type

285. Find the general solution of the differential equation y dx – (x + 2 y²) dy = 0.

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286. Solve the following differential equation:
(1 + y²)dx = (tan^{– 1} y – x) dy

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287. Solve the following differential equation:

square root of 1 minus straight y squared end root space dx equals left parenthesis sin to the power of negative 1 end exponent straight y space minus straight x right parenthesis space dy

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Short Answer Type

288. Find a one parameter family of solutions of each of the following differential equation e^–y sec² y dy = dx + x dy.

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Long Answer Type

289. Find the particular solution of the differential equation:

dy over dx plus straight y space cotx space equals space 2 straight x plus straight x squared space cotx space space space left parenthesis straight x not equal to 0 right parenthesis

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290. For the given differential equation, find a particular solution satisfying the given condition:

dy over dx plus 2 straight y space tan space straight x space equals space sin space straight x space colon space straight y space space equals space 0 space space when space straight x space equals space straight pi over 3

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Previous Year Papers

Test Series

Test Yourself

Long Answer Type

Short Answer Type

282. Solve:

Long Answer Type

Short Answer Type

Long Answer Type

Short Answer Type

Long Answer Type

282.
Solve: $open parentheses straight x plus straight y close parentheses space dy over dx space equals space 1.$