Find the particular solution of the differential equation(1 + e2

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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.


The given differential equation is
              (1 + e2x ) dy + (1 + y2 ) ex dx = 0   or   (1 + e2x ) dy  = - (1 + y2 ) ex dx

therefore space space space space fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals negative space fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx
rightwards double arrow space space space space integral fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space minus integral fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx 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 space space space space space space space space space... left parenthesis 1 right parenthesis
Let I = integral fraction numerator straight e to the power of straight x over denominator 1 plus straight e to the power of 2 straight x end exponent end fraction dx

Put         straight e to the power of straight x space equals space straight t comma space space space therefore space space space space straight e to the power of straight x space dx space equals space dt
therefore space space space space straight I space equals space integral fraction numerator dt over denominator 1 plus straight t squared end fraction space equals space tan to the power of negative 1 end exponent straight t space equals space tan to the power of negative 1 end exponent left parenthesis straight e to the power of straight x right parenthesis
therefore space space space space from space left parenthesis 2 right parenthesis comma space space tan to the power of negative 1 end exponent straight y space equals space tan left parenthesis straight e to the power of straight x right parenthesis space plus straight c 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 2 right parenthesis
Now,   x = 0,  y = 1
therefore space space space tan to the power of negative 1 end exponent 1 space equals space minus tan left parenthesis straight e to the power of 0 right parenthesis space plus space straight c space space space rightwards double arrow space space space space straight pi over 4 space equals space minus tan to the power of negative 1 end exponent left parenthesis 1 right parenthesis space plus space straight c
rightwards double arrow space space space straight pi over 4 space equals space minus straight pi over 4 plus straight c space space space rightwards double arrow space space space space straight c space equals space straight pi over 2
therefore space space from space left parenthesis 2 right parenthesis comma space tan to the power of negative 1 end exponent straight y equals space minus tan space straight e to the power of straight x plus straight pi over 2 comma space which space is space required space solution. space

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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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