Find the solution of the equation:            subject to

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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.


dy over dx space equals space straight e to the power of straight x plus straight y end exponent plus straight x squared straight e to the power of straight x space space space space space space rightwards double arrow space space space dy over dx space equals space straight e to the power of straight x. space straight e to the power of straight y space plus space straight x squared space straight e to the power of straight x space space rightwards double arrow space space dy over dx space equals space left parenthesis straight e to the power of straight x plus straight x squared right parenthesis space straight e to the power of straight y
Separating the variables, we get,  dy over straight e to the power of straight y space equals space left parenthesis straight e to the power of straight x plus straight x squared right parenthesis space dx
therefore space space space space space space space integral straight e to the power of negative straight y end exponent dy space equals space integral left parenthesis straight e to the power of straight x plus straight x squared right parenthesis space dx space space space space space rightwards double arrow space space space fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction space equals space straight e to the power of straight x plus straight x cubed over 3 plus straight c
or             straight e to the power of negative straight y end exponent plus straight e to the power of straight x plus straight x cubed over 3 plus straight c space equals space 0                       ...(1)
When x = 0,  y = 0,  we have
                straight e to the power of 0 plus straight e to the power of 0 plus 0 plus straight c space equals space 0 space space space space rightwards double arrow space space space space 1 plus 1 plus straight c space equals space 0 space space space rightwards double arrow space space space straight c space equals negative space 2
Putting c = -2 in (1), the required solution is
                     e to the power of italic minus x end exponent italic plus e to the power of x italic plus x to the power of italic 3 over italic 3 italic minus italic 2 italic space italic equals italic space italic 0 italic.
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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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