Find the equation of a curve passing through the point (– 2, 3

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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

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

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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.


We know that slope of tangent to a curve is given by dy over dx
From the given condition, 
                       dy over dx space equals space fraction numerator 2 straight x over denominator straight y squared end fraction
Separating the variables and integrating, 
                   integral straight y squared space dy space equals space 2 space integral space straight x space dx
therefore space space space space space space space space space space space space space straight y cubed over 3 space equals space straight x squared plus straight c                             ...(1)
Since curve passes through (-2, 3)
therefore space space space space space space space space space space space space space space space space fraction numerator left parenthesis 3 right parenthesis cubed over denominator 3 end fraction space equals space left parenthesis negative 2 right parenthesis squared plus straight c space space space space or space space space space space space 9 space equals space 4 plus straight c space space space rightwards double arrow space space space space straight c space equals space 5
Putting c = 5 in (1), we get,
            straight y cubed over 3 space equals space straight x squared plus 5 space space space or space space space straight y cubed space equals space 3 left parenthesis straight x squared plus 5 right parenthesis
which is required equation of curve. 
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