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Integrating factor of differential equat...

Integrating factor of differential equation `cosx(dy)/(dx)+ysinx=1` is (a) `( b ) (c)cosx (d)` (e) (b) `( f ) (g)tanx (h)` (i) (c) `( d ) (e)secx (f)` (g) (d) `( h ) (i)sinx (j)` (k)

A

`cosx`

B

`tanx`

C

`secx`

D

`sinx`

Text Solution

Verified by Experts

The correct Answer is:
C

`cosx(dy)/(dx)+ysinx=1`
or `(dy)/(dx)+y(sinx)/(cosx)=secx`
`therefore intPdx=int(sinx)/(cosx)dx`
`=-logcosx`
`=log secx`
`therefore` I.F. `=e^(intPdx)=e^(log secx)=secx`
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CENGAGE-DIFFERENTIAL EQUATIONS-Exercise (Single)
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  2. The solution of the differential equation x(x^(2)+1)(dy//dx)=y(1-x^(2)...

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  3. Integrating factor of differential equation cosx(dy)/(dx)+ysinx=1 is (...

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  4. Solution of the equation cos^2x(dy)/(dx)-(tan2x)y=cos^4x ,|x|<pi/4, wh...

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  5. If integrating factor of x(1-x^2)dy+(2x^2y-y-a x^3)dx=0 is e^(intp dx...

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  6. A function y=f(x) satisfies (x+1)f^(prime)(x)-2(x^2+x)f(x)=(e^x^2)/((x...

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  7. The general solution of the equation (dy)/(dx)=1+x y is (a) ( b ) (...

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  8. The solution of the differential equation ((x+2y^3)dy)/(dx)=y is (a) ...

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  9. The solution of the differential equation x^2(dy)/(dx)cos1/x-ysin1/x=-...

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  10. The solution of (dy)/(dx)=(x^2+y^2+1)/(2x y) satisfying y(1)=1 is give...

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  11. The solution of the differential equation (dy)/(dx) = 1/(xy[x^(2)sin...

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  12. The equation of a curve passing through (2,7/2) and having gradient...

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  13. Which of the following is not the differential equation of family of c...

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  14. Tangent to a curve intercepts the y-axis at a point Pdot A line ...

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  15. Orthogonal trajectories of family of the curve x^(2/3)+y^2/3=a^((2/3))...

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  16. The curve in the first quadrant for which the normal at any point (...

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  17. The equation of the curve which is such that the portion of the axis o...

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  18. The family of curves represented by (dy)/(dx)=(x^(2)+x+1)/(y^(2)+y+1) ...

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  19. A normal at P(x , y) on a curve meets the x-axis at Q and N is the ...

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  20. A curve is such that the mid-point of the portion of the tangent in...

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