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.

Differential Equations

Short Answer Type

141.

Solve:
$open parentheses straight e to the power of straight x plus 1 close parentheses space straight y space dy space plus space left parenthesis straight y plus 1 right parenthesis space straight e to the power of straight x space dx space equals space 0$

87 Views

142.

Solve:
$straight y left parenthesis 1 minus straight x squared right parenthesis space dy space plus space straight x space left parenthesis 1 plus straight y squared right parenthesis space dx space equals space 0.$

83 Views

143.

Solve:
$straight y space logydx space minus space straight x space dy space equals space 0$

91 Views

Long Answer Type

144.

Show that the general solution of the differential equation $dy over dx plus fraction numerator straight y squared plus straight y plus 1 over denominator straight x squared plus straight x plus 1 end fraction space equals 0$ is given by (x + y + 1) = A (1 – x – y – 2 x y), where A is parameter.

77 Views

Short Answer Type

145.

Solve:
$dy over dx space equals space fraction numerator straight x left parenthesis 2 space log space straight x space plus space 1 right parenthesis over denominator sin space straight y space plus space straight y space cosy end fraction$

80 Views

146.

Solve
$dy over dx space equals space fraction numerator xe to the power of straight x logx plus straight e to the power of straight x over denominator straight x space cosy end fraction$

83 Views

Long Answer Type

147.

Solve
$dy over dx space equals space sin cubed straight x space cos squared straight x plus straight x space straight e to the power of straight x$

82 Views

148.
Solve:
$dy over dx space equals space cos cubed straight x space sin to the power of 4 straight x plus straight x square root of 2 straight x plus 1 end root$

The given differential equation is

dy over dx space equals space cos cubed straight x space sin to the power of 4 straight x plus straight x square root of 2 straight x plus 1 end root

Separating the variables, we get,

dy space equals space left parenthesis cos cubed straight x space sin to the power of 4 straight x plus straight x square root of 2 straight x plus 1 end root right parenthesis space dx

therefore space space space integral dy space equals space integral left parenthesis cos cubed straight x space sin to the power of 4 straight x plus straight x square root of 2 straight x plus 1 end root right parenthesis space dx

or space space space integral dy space equals space integral cos cubed straight x space sin to the power of 4 straight x space dx plus integral straight x space square root of 2 straight x plus 1 end root space dx

...(1)
Let I =

integral space cos cubed straight x space sin to the power of 4 straight x space dx space equals space integral cos squared straight x. space sin to the power of 4 straight x. space cosx space dx

equals space integral left parenthesis 1 minus sin squared straight x right parenthesis space sin to the power of 4 straight x. space cosx space dx

Put

sin space straight x space equals space straight t comma space space space therefore space space cos space straight x space dx space equals space dt

therefore space space space straight I space equals space integral left parenthesis 1 minus straight t squared right parenthesis space straight t to the power of 4 space dt space equals space integral left parenthesis straight t to the power of 4 minus straight t to the power of 6 right parenthesis space dt space equals space 1 fifth space straight t to the power of 5 space minus space 1 over 7 straight t to the power of 7 space equals 1 fifth sin to the power of 5 straight x minus 1 over 7 sin to the power of 7 straight x

Let

straight I subscript 1 space equals space integral straight x square root of 2 straight x plus 1 end root space dx

Put

square root of 2 straight x plus 1 end root space equals space straight y comma space space space space space therefore space space 2 straight x plus 1 space equals space straight y squared space space rightwards double arrow space space space 2 straight x space equals space straight y squared minus 1 space space rightwards double arrow space straight x space equals space 1 half left parenthesis straight y squared minus 1 right parenthesis

therefore space space space space dx space equals space 1 half space 2 straight y space dy space equals space straight y space dy therefore space space space straight I subscript 1 space equals space integral fraction numerator straight y squared minus 1 over denominator 2 end fraction. straight y. space straight y space dy space equals space 1 half integral left parenthesis straight y squared minus 1 right parenthesis space straight y squared space dy space equals space 1 half integral left parenthesis straight y to the power of 4 minus straight y squared right parenthesis space dy space equals 1 half open square brackets straight y to the power of 5 over 5 minus straight y cubed over 3 close square brackets space space space space space space space space space space space space equals space 1 over 10 straight y to the power of 5 space minus space 1 over 6 straight y cubed space equals space 1 over 10 left parenthesis 2 straight x plus 1 right parenthesis to the power of 5 divided by 2 end exponent minus space 1 over 6 left parenthesis 2 straight x plus 1 right parenthesis to the power of 3 divided by 2 end exponent

therefore space space space space from space left parenthesis 1 right parenthesis comma space we space get comma space space space space space space space space space space space integral dy space equals space 1 fifth sin to the power of 5 straight x minus 1 over 7 sin to the power of 7 straight x plus 1 over 10 left parenthesis 2 straight x plus 1 right parenthesis to the power of 5 divided by 2 end exponent minus 1 over 6 left parenthesis 2 straight x plus 1 right parenthesis to the power of 3 divided by 2 end exponent therefore space space space space straight y space equals space 1 fifth sin to the power of 5 straight x minus 1 over 7 sin to the power of 7 straight x plus 1 over 10 left parenthesis 2 straight x plus 1 right parenthesis to the power of 5 divided by 2 end exponent minus 1 over 6 left parenthesis 2 straight x plus 1 right parenthesis to the power of 3 divided by 2 end exponent plus straight c

84 Views

149.

Solve:
$dy over dx equals negative straight x space sin squared straight x space space equals space fraction numerator 1 over denominator straight x space log space straight x end fraction$