Find the equation of the curve passing through the point (1, 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.

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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).


The given differential equation is
                         x dy = (2 x2 + 1) dx
or               dy space equals space fraction numerator 2 straight x squared plus 1 over denominator straight x end fraction dx
Integrating, integral space 1 space dy space equals space integral fraction numerator 2 straight x squared plus 1 over denominator straight x end fraction dx
or               integral 1 space dy space equals space integral open parentheses fraction numerator 2 straight x squared over denominator straight x end fraction plus 1 over straight x close parentheses dx
or                integral 1 space dy space equals space integral open parentheses 2 straight x plus 1 over straight x close parentheses dx
therefore 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 straight y space equals space straight x squared space plus log space open vertical bar straight x close vertical bar space plus straight c                     ...(1)
The curve passes through (1, 1)
therefore space space space 1 space equals space 1 plus log space open vertical bar 1 close vertical bar plus straight c space space space or space space space space 1 space equals space 1 space plus 0 plus straight c space space space space rightwards double arrow space space space straight c space space equals 0
Putting c = 0 in (1), we get
             straight y space equals straight x squared plus space log space open vertical bar straight x close vertical bar comma space which space is space required space solution. space
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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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