Solve the following initial value problem:(1 + x y) y dx + (1 �

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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.


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

or            left parenthesis straight y space dx space plus space straight x space dy right parenthesis space plus space straight x space straight y squared space dx space minus space straight x squared straight y space dy space equals space 0

or            fraction numerator ydx plus xdy over denominator straight x squared straight y squared end fraction plus 1 over straight x dx minus 1 over straight y dy space equals space 0

therefore space space space space space integral fraction numerator straight y space dx space plus space straight x space dy over denominator straight x squared straight y squared end fraction plus integral 1 over straight x dx space minus space integral 1 over straight y dy space equals space 0               ...(1)

Let I = integral fraction numerator straight y space dx space plus space straight x space dy over denominator straight x squared straight y squared end fraction
Put x y  = t so that  x dy + y dx = dt
therefore space space space space straight I space equals space integral 1 over straight t squared dt space equals space integral straight t to the power of negative 2 end exponent dt space equals space fraction numerator straight t to the power of negative 1 end exponent over denominator negative 1 end fraction space equals space minus 1 over straight t space equals negative fraction numerator 1 over denominator straight x space straight y end fraction
therefore  from (1),
        negative 1 over xy plus logx minus logy space equals space straight c                             
Now     straight y left parenthesis 1 right parenthesis space equals space 1 space space space space space space space space space space space space space space space rightwards double arrow space space space space straight y space equals 1 space space space space when space straight x space equals space 1
therefore space space space minus 1 over 1 plus log space 1 space minus space log space 1 space equals space straight c space space space rightwards double arrow space space space space straight c space equals space minus 1
therefore space space space space from space left parenthesis 1 right parenthesis comma space solution space of space differential space equation space is space
space space space space space space space space space minus 1 over xy plus log space straight x space minus space log space straight y space equals space minus 1
therefore space space space log space straight x space equals space log space straight y space plus space fraction numerator 1 over denominator straight x space straight y end fraction minus 1
which is required solution. 
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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

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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