Find the particular solution of the differential equation  giv

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 Multiple Choice QuestionsShort Answer Type

151.

Solve the differential equation:
dy over dx space equals space straight y space sin space 2 straight x comma space space space given space that space space straight y left parenthesis 0 right parenthesis space equals space 1.

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152.

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

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153. Find the particular solution of the differential equation dy over dx space equals space minus 4 xy squared given that y = 1,  when x = 0.

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154. Find the particular solution of the differential equation
(1 + e2x ) dy + (1 + y2 ) ex dx = 0, given that y = 1 when x = 0.
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155. Find the solution of the equation:
            dy over dx space equals space straight e to the power of straight x plus straight y end exponent plus straight x squared space straight e to the power of straight x
subject to the condition, when x = 0;  y = 0.
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156.

Solve the differential equation;
x (1 + y2 ) dx – y (1 + x2 ) dy = 0  given that y = 0 when x = 1. 

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 Multiple Choice QuestionsLong Answer Type

157. Find the particular solution of (1 + x2 + y2 + x2 y2 ) dx + x y dy = 0
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 Multiple Choice QuestionsShort Answer Type

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158. Find the particular solution of the differential equation log space open parentheses dy over dx close parentheses space equals space 3 straight x plus 4 straight y given that y = 0 when x = 0.


The given differential equation is
log open parentheses dy over dx close parentheses space equals space 3 straight x plus 4 straight y space space space space or space space space space space dy over dx equals space straight e to the power of 3 straight x plus 4 straight y end exponent space space space or space space space space space dy over dx space equals straight e to the power of 2 straight x end exponent. space straight e to the power of 4 straight y end exponent
Separating the variables, dy over straight e to the power of 4 straight y end exponent space equals straight e to the power of 3 straight x end exponent space dx space or space space space integral straight e to the power of negative 4 straight y end exponent dy space equals space integral straight e to the power of 3 straight x end exponent space dx
therefore space space space fraction numerator straight e to the power of negative 4 straight y end exponent over denominator negative 4 end fraction space equals space straight e to the power of 3 straight x end exponent over 3 plus straight c                                    ...(1)
when y = 0,  x = 0,   then from (1), we get,
                negative straight e to the power of 0 over 4 space equals space straight e to the power of 0 over 3 plus straight c space space space space or space space space minus 1 fourth space equals space 1 third plus straight c
therefore space space space space space space straight c space equals space minus 1 fourth minus 1 third space equals space minus 7 over 12
therefore space space from (1),  1 fourth straight e to the power of negative 4 straight y end exponent space equals space 1 third straight e to the power of 3 straight x end exponent space minus space 7 over 12 which  is required solution. 

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159.

Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2 x2 + 1) dx (x ≠ 0).

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 Multiple Choice QuestionsLong Answer Type

160. Find the equation of a curve passing through the point (– 2, 3), given that the slope of the tangent to the curve at any point (x, y) is fraction numerator 2 straight y over denominator straight y squared end fraction.
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