Find the particular solution of (1 + x2 + y2 + x2 y2 ) dx +

Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.
Advertisement

 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.

87 Views

152.

Solve the following initial value problem:
(1 + x y) y dx + (1 – x y) x dy = 0, y (1) = 1.

114 Views

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.

88 Views

154. Find the particular solution of the differential equation
(1 + e2x ) dy + (1 + y2 ) ex dx = 0, given that y = 1 when x = 0.
91 Views

Advertisement
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.
88 Views

156.

Solve the differential equation;
x (1 + y2 ) dx – y (1 + x2 ) dy = 0  given that y = 0 when x = 1. 

84 Views

 Multiple Choice QuestionsLong Answer Type

Advertisement

157. Find the particular solution of (1 + x2 + y2 + x2 y2 ) dx + x y dy = 0


The given differential equation is
 (1 + x2 + y2 + x2 y2 ) dx + x y dy = 0
or   open square brackets left parenthesis 1 plus straight x squared right parenthesis space plus space straight y squared left parenthesis 1 plus straight x squared right parenthesis close square brackets dx space plus space straight x space straight y space dy space equals space 0
or   left parenthesis 1 plus straight x squared right parenthesis space left parenthesis 1 plus straight y squared right parenthesis space dx space space plus straight x space straight y space dy space equals space 0
or                      straight x space straight y space dy space equals space minus left parenthesis 1 plus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis space dx
therefore space space space space space fraction numerator straight y over denominator 1 plus straight y squared end fraction dy space equals space minus fraction numerator 1 plus straight x squared over denominator straight x end fraction dx
therefore space space space space fraction numerator straight y over denominator 1 plus straight y squared end fraction dy space equals space minus open parentheses 1 over straight x plus straight x close parentheses dx
Integrating,   1 half integral fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy space equals space minus integral open parentheses 1 over straight x plus straight x close parentheses space dx
therefore space space space space 1 half space log space left parenthesis 1 plus straight y squared right parenthesis space equals space minus open square brackets log space open vertical bar straight x close vertical bar plus straight x squared over 2 close square brackets plus straight c                   ...(1)
Now straight y space equals space 0 space space when space straight x space equals space 1
therefore space space space space space space space space space space space space space space space 1 half log space 1 space space equals space minus open square brackets log space 1 plus 1 half close square brackets plus straight c
therefore space space space space space space space space space space space space space space space space 0 space equals space minus 1 half plus straight c space space space space space space space space space space space space space space rightwards double arrow space space space straight c space equals space 1 half
therefore space space space space from space left parenthesis 1 right parenthesis
space space space space space space space space space space space space space 1 half log space open square brackets 1 plus straight y squared close square brackets equals space minus open square brackets log space open vertical bar straight x close vertical bar plus straight x squared over 2 close square brackets plus 1 half
which is required solution. 
87 Views

Advertisement

 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.
77 Views

Advertisement
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).

78 Views

 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.
82 Views

Advertisement