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**Q.1. If y = y (x) is the solution of the differential equationsuch that y (0) = 0, then y (1) is equal to (2020)****(1) 1 + log _{e}2**

Ans.

We have,

Let e

Now, the differential equation becomes

Integrating factor of equation is

So, the solution of differential equation is

Now, at x = 0, y = 0, then c =1. Therefore,

Substituting x =1, we get y(1) = 1 + log

(1) 2 â€“ e

(2) âˆ’e

(3) 2

(4) 2 + e

Ans.

Solution.

The given differential equation is

This is a linear differential equation.

Therefore,

Now, the solution of differential equation is given by

Given: y (0) =1

So, 0 = e + c â‡’ c = -e.

Therefore,

Hence, the abscissa of point of intersection with the x-axis is

x = (0 âˆ’ 0 + 2) - e = 2 - e

(1)

(2)

(3)

(4)

Ans.

Solution.

We have

Integrating both sides of equation, we get sin

Now,

sin

At x = -1/âˆš2 , we get y = sin

(1)

(2)

(3) xy'' = y'

(4)

Ans.

Solution.

The equation of the family of curve is x

Differentiating Eq. (1) w.r.t. x, we get

Substituting the value of b in Eq. (1), we get

Ans.

Solution.

We have

.....(1)

This is a linear differential equation of first order. So,

Therefore, the solution of differential equation is

Given, y (2) = 0 â‡’ c + 2 + 1 = 0 â‡’ c = -3.

Therefore,

Substituting x = 3 in above equation, we get

(1)

(2)

(3)

(4)

Ans.

Solution.

We have..... (1)

Substitutingin Eq. (1), we get

Integrating both sides of the equation, we get

Now, y(1) = 1 â‡’

So,

For y(x) = e, we have**Q.7. If y = y(x) is the solution of the differential equation, satisfying y(1) = 1, thenis equal to: (2019)(1) 7/64(2) 1/2(3) 49/16(4) 13/16**

Solution of differential equation is:

...(1)

âˆµ y(1) = 1

âˆ´ C = 3/4

Then, from equation (1)

âˆ´

âˆ´

f(xy) = f(x).f(y) ...(1)

By first principle derivative formula,

Solution.

y

2xydy - y

d(xy

...(1)

Since, the above curve passes through the point (1,1)

Now, the curve (1) becomes

y

â‡’ y

(x - 1)

The above equation represents a circle with centre (1, 0) and centre lies on x-axis.

** (2019)****Ans.** (3)**Solution. **Given differential equation is,

Complete solution is given by

Differentiate both sides with respect to x,

...(1)

Now, from equation (1)

**Q.14. Let y = y(x) be the solution of the differential equation, such that y(0) = 0. If , then the value of 'a' is: (2019)****(1) 1/4****(2) 1/2****(3) 1****(4) 1/16****Ans. **(4)**Solution.**

Since, the above differential equation is a linear differential equation

Then, the solution of the differential equation

If x = 0 then y = 0 (given)

â‡’ 0 = 0 + c

â‡’ c = 0

Then, equation (1) becomes,

â‡’ y (1 + x^{2}) = tan^{-1} x

Now put x = 1 in above equation, then**Q.15. The solution of the differential equation (x â‰ 0) with y(1) = 1, is: (2019)****Ans. **(3)**Solution.**

Since, the above differential equation is the linear differential equation, then

Now, the solution of the linear differential equation**Q.16. and then is equal to: (2019)****Ans. **(3)

Solution.

Now, put y = Ï€/6 in the above equation,

Q.17. If y = y(x) is the solution of the differential equation dy/dx = (tan x - y)sec^{2} x, such that y (0) = 0, then (2019)**(1) e - 2****Ans. **(1)

Solution.

Given equation is linear differential equation.

Hence, solution of differential equation,**Q.18. Let y = y(x) be the solution of the differential equation, such that y (0) = 1. (2019)****Then:****Ans. (4)****Solution.**

Given differential equation is.**Q.19. Consider the differential equation, value of y is 1 when x = 1, then the value of x for which y = 2, is: (2019)****Ans. **(2)**Solution.**

Consider the differential equation,**Q.20. ****The general solution of the differential equation (y ^{2} - x^{3}) dx - xydy = 0 (x â‰ 0) is : (2019)**

(2) y

(3) y

(4) y

Given differential equation can be written as,

(2018)

(1)

(2)

(3)

(4)

Ans.

Solution:

Integrating both sides we get y sin x = 2x^{2}+ C**Q.22. Let y = y(x) be the solution of the differential equation where f(x) . If y(0) = 0, then y(3/2) is: (2018)****(1) 1/2e****(2) ****(3) ****(4) ****Ans.****Solution. **

dy/dx + 2y = f(x) is a linear differential equation

solution of the above equation is**Q.23. ****The differential equation representing the family of ellipses having foci either on the x-axis or on the y-axis, centre at the origin and passing through the point (0, 3) is: (2018)(1) xyy' - y ^{2} + 9 = 0 (2) xyy" + x (y')^{2} - yy' = 0(3) xyy' + y^{2} - 9 = 0(4) x + yy" = 0Ans.**(2)

Solution.

Equation of ellipse

since it passes through (0, 3)â‡’b

âˆ´ Equation of ellipse becomes

â‡’

......(2)

(1) 4/3

(2) 1/3

(3) -2/3

(4) -1/3

Ans.

ln| y + 1 |+ ln (2 + sinx ) = lnC

(y + 1) (2+ sinx ) = C

Put x = 0, y = 1

(1 + 1)Ã— 2 = C â‡’ C = 4

Now, ( y +1) (2+ sinx ) = 4

( y +1) (2+ 1) = 4

(1)

(2)

(3)

(4)

Ans.

ydx - xdy - 3y

Passes through (1, 1) âˆ´ 1 = 3 + c ; c = -2

x = 3y

(1) 26

(2) -24

(3) -23

(4) -26

Ans.

(1) -2/5

(2) -4/5

(3) 2/5

(4) 4/5

Ans.

We have to find

Put x = -1/2

(1)

(2)

(3)

(4)

Ans.

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