Solve the differential equation;x (1 + y2 ) dx – y (1 + x2 )

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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. 


The given differential equation is
           x (1 + y2 ) dx – y (1 + x2 ) dy = 0 
or              y (1 + x2 ) dy = 0 =    x (1 + y2 ) dx

therefore space space space space space space space space space space space space fraction numerator straight y over denominator 1 plus straight y squared end fraction dy space equals space fraction numerator straight x over denominator 1 plus straight x squared end fraction dx
rightwards double arrow space space space space space space space space space space fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy space equals space fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction dx

Integrating, integral fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy space equals space integral fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction dx
therefore space space space space space space space space space space space space space log space left parenthesis 1 plus straight y squared right parenthesis space equals space log space left parenthesis 1 plus straight x squared right parenthesis space plus space log space straight c
therefore space space space space space space space space log space left parenthesis 1 plus straight y squared right parenthesis space equals space log space left square bracket straight c space left parenthesis 1 plus straight x squared right parenthesis right square bracket
rightwards double arrow space space space space space space space space space space space space space space space space space space space space space space space space 1 plus straight y squared space equals space straight c space left parenthesis 1 plus straight x squared right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Now   y = 0 when x = 1
therefore space space space space space space space space space 1 plus 0 space equals space straight c space left parenthesis 1 plus 1 right parenthesis space space or space space space space 1 space equals space 2 straight c space space space space rightwards double arrow space space straight c space equals space 1 half
therefore space space space space from space left parenthesis 1 right parenthesis comma space space space space space space space 1 plus straight y squared space equals space 1 half left parenthesis 1 plus straight x squared right parenthesis
which is required solution. 
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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

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.
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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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