Tuesday, January 31, 2017

linear differential equation for cbse ncert miscellaneous problem

linear differential equation for cbse ncert miscellaneous problem

solve [ { e^[-2sqrt(x)] / sqrt(x) } - {y / sqrt(x)}] [dx/dy] = 1  [x is not 0]

here there is only one term in y

so we have to rearrange the equation in the form

[dy/dx] + Py = Q

identify the values of P and Q

find the integrating factor e^[integral of P]

and use in the solution

y[integrating factor] = integral of [Q*integrating factor] + C









Variable separable differential equation
*find the equation of the curve passing through (0,pi/4) whose differential equation is
sinx cosy dx +  cosx siny dy =0
solution of  variable separable differential equation find the equation of the curve passing through (0,pi/4) whose differential equation issinx cosy dx +  cosx siny dy =0

* show that the general solution of y' + {[ (y^2)+y + 1]/[x^2+x+1] = 0
is given by x+y+1=A[1-x-y-2xy]

*solve y e^(x/y) = [ y e^(x/y) + (y^2) ]dy  [ y is not equal to 0]
solution of differential equation which is not homogeneous but which can be solved using x=vy



solution of a second order differential equation using reduction of order
solve y"-y = 0 if y = coshx is one of the solutions
using the formula for reduction of order
solution of solution of a second order differential equation using reduction of order

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on  orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

find the orthogonal trajectory of y = (k/x)
solution to  find the orthogonal trajectory of y = (k/x)



formulae on integration
 
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PAGE 2 INTEGRATION BY SUBSTITUTION

 PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR


PAGE 6 INTEGRATION USING PARTIAL FRACTIONS

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non homogeneous equation with the substitution x=vy

miscellaneous problem from ncert differential equation where the substitution x=vy still works even though is not a homogeneous equation

solve y e^(x/y) = [ y e^(x/y) + (y^2) ]dy  [ y is not equal to 0]

here there  are terms containing [x/y]
so extract dx/dy call the value of dx/dy as f[x,y]
replace x with tx and y with ty and check if the function is homogeneous

here f[tx,ty] is not equal to f[x,y]
therefore the function is not homogeneous

but the substitution v =(x/y) will still work for this problem









Variable separable differential equation


* show that the general solution of y' + {[ (y^2)+y + 1]/[x^2+x+1] = 0
is given by x+y+1=A[1-x-y-2xy]

solution of a second order differential equation using reduction of order
solve y"-y = 0 if y = coshx is one of the solutions
using the formula for reduction of order
solution of solution of a second order differential equation using reduction of order

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on  orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

find the orthogonal trajectory of y = (k/x)
solution to  find the orthogonal trajectory of y = (k/x)



formulae on integration
 
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PAGE 2 INTEGRATION BY SUBSTITUTION

 PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR


PAGE 6 INTEGRATION USING PARTIAL FRACTIONS

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miscellaneous problem from differential equations

miscellaneous problem from differential equations
find the equation of the curve passing through (0,pi/4) whose differential equation is
sinx cosy dx +  cosx siny dy =0

here cosy with dx and cosx with dy should be removed.

So divide each term with cosy cosx

the resulting equation is of variable separable type.

 integrate term by term

Now to get the value of C, use the given condition that the curve passes through  (0,pi/4)

put x =0  and y = pi/4 and solve for C.





Variable separable differential equation
* show that the general solution of y' + {[ (y^2)+y + 1]/[x^2+x+1] = 0
is given by x+y+1=A[1-x-y-2xy]


*solve y e^(x/y) = [ y e^(x/y) + (y^2) ]dy  [ y is not equal to 0]
solution of differential equation which is not homogeneous but which can be solved using x=vy

solution of a second order differential equation using reduction of order
solve y"-y = 0 if y = coshx is one of the solutions
using the formula for reduction of order
solution of solution of a second order differential equation using reduction of order

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on  orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

find the orthogonal trajectory of y = (k/x)
solution to  find the orthogonal trajectory of y = (k/x)



formulae on integration
 
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PAGE 2 INTEGRATION BY SUBSTITUTION

 PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR


PAGE 6 INTEGRATION USING PARTIAL FRACTIONS

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variable separable differential equation for cbse ncert 12th

variable separable differential equation for cbse ncert 12th

show that the general solution of y' + {[ (y^2)+y + 1]/[x^2+x+1] = 0
is given by x+y+1=A[1-x-y-2xy]

first separate the variables , then, complete the squares in each term,  integrate term by term
then use the formula for arctanx+arctany

 








homogeneous differential equation
prove that [ x^2 - y^2 ]=c [ x^2 - y^2 ]^2 is a solution of
[ x^3 - 3x (y^2) ] dx = [ y^3 - 3x^2y ] dy

solution of homogeneous differential equation from miscellaneous problems of ncert cbse 12th mathematics

solution of a second order differential equation using reduction of order
solve y"-y = 0 if y = coshx is one of the solutions
using the formula for reduction of order
solution of solution of a second order differential equation using reduction of order

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on  orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

find the orthogonal trajectory of y = (k/x)
solution to  find the orthogonal trajectory of y = (k/x)



formulae on integration
 
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PAGE 2 INTEGRATION BY SUBSTITUTION

 PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR


PAGE 6 INTEGRATION USING PARTIAL FRACTIONS

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Thursday, January 26, 2017

solution of homogeneous differential equation

solution of homogeneous differential equation

prove that [ x^2 - y^2 ]=c [ x^2 - y^2 ]^2 is a solution of
[ x^3 - 3x (y^2) ] dx = [ y^3 - 3x^2y ] dy

first solve for (dy/dx)
take the function as f(x,y)
now check if the equation is a homogeneous differential equation
replace x and y with tx and ty
simplify and see if the t can be cancelled off to get back f(x,y)
If the equation is a homogeneous differential equation, put y=Vx
(dy/dx) = V + x(dV/dx)
simplify and separate the variables and then integrate









Variable separable differential equation
*find the equation of the curve passing through (0,pi/4) whose differential equation is
sinx cosy dx +  cosx siny dy =0
solution of  variable separable differential equation find the equation of the curve passing through (0,pi/4) whose differential equation issinx cosy dx +  cosx siny dy =0

* show that the general solution of y' + {[ (y^2)+y + 1]/[x^2+x+1] = 0
is given by x+y+1=A[1-x-y-2xy]

*solve y e^(x/y) = [ y e^(x/y) + (y^2) ]dy  [ y is not equal to 0]
solution of differential equation which is not homogeneous but which can be solved using x=vy

*linear differential equation 
solve [ { e^[-2sqrt(x)] / sqrt(x) } - {y / sqrt(x)}] [dx/dy] = 1  [x is not 0]
solution of linear differential equation of ncert cbse miscellaneous differential equation

*linear differential equation of the type
solve y dx + [x-(y^2)]dy = 0
solution of linear differential equation of the type [dx/dy] + Px = Q


*find a particular solution of the differential equation [dy/dx]+ycotx =4xcosecx [x is not 0] given that y =0 if x=pi/2
solution of linear differential equation from ncert cbse miscellaneous  [dy/dx]+ycotx =4xcosecx [x is not 0] given that y =0 if x=pi/2

solution of a second order differential equation using reduction of order
solve y"-y = 0 if y = coshx is one of the solutions
using the formula for reduction of order
solution of solution of a second order differential equation using reduction of order

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on  orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

find the orthogonal trajectory of y = (k/x)
solution to  find the orthogonal trajectory of y = (k/x)



formulae on integration
 
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PAGE 2 INTEGRATION BY SUBSTITUTION

 PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR


PAGE 6 INTEGRATION USING PARTIAL FRACTIONS

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form the differential equation of the family of circles in the first quadrant which touch the coordinate axes

form the differential equation of the family of circles in the first quadrant which touch the coordinate axes

Take the radius as r so that the centre is (r,r)
write the equation of the circle in terms of r
r is the aribitrary constant.
so differentiate it one time with respect to x and use
that to solve for r
substitute into (x-r) and (y-r) and substitute into the equation of the
circle to eliminate r





solution of a second order differential equation using reduction of order
solve y"-y = 0 if y = coshx is one of the solutions
using the formula for reduction of order
solution of solution of a second order differential equation using reduction of order

variation of parameter method

solve xy" - 4y' = x^4 by method of variation of parameter
solution to problem on differential questions using variation of parameter method

orthogonal trajectory of y(1+x ² ) = Cx

find the orthogonal trajectory of y(1+x ² ) = Cx
answer to problem on  orthogonal trajectory of y(1+x ² ) = Cx

orthogonal trajectory of y = (k/x)

find the orthogonal trajectory of y = (k/x)
solution to  find the orthogonal trajectory of y = (k/x)



formulae on integration
 
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PAGE 2 INTEGRATION BY SUBSTITUTION

 PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

PAGE 4 INTEGRATION BY PARTS

PAGE 5 INTEGRATION BY MANIPULATION OF NUMERATOR IN TERMS OF DENOMINATOR


PAGE 6 INTEGRATION USING PARTIAL FRACTIONS

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Tuesday, January 24, 2017

integral of e^(2-3x) from 0 to 1 using limit of sums

integral of e^(2-3x) with limit  0 to 1 using limit of sums

identify a =0 , b=1

nh = b-a = 1

f(x)  = e^(2-3x)

find f(a) = f(0)
f(a+h)=f(h)
f(a+2h) = f[2h]
till the pattern can be identified,
f(a + (n-1) h ) = f[(n-1) h] etc

simplify using properties of sum of  n terms of a GP.
and use the limit [(e^h) -1] / h tends to 1 as h tends to 0





formulae on integration
 
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 PAGE 3 INTEGRATION BY COMPLETION OF SQUARES

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integral of 2*cube of (tanx) with 0 to pi/4 as limits

integral of 2* (tanx)^3 with 0 to pi/4 as limits

write (tanx)^3 as the product of tanx and (tanx)^2

use  trigonometric formulae to rewrite  (tanx)^2 as [ (secx)^2 -1]

multiply the  tanx into [ (secx)^2 -1] and split into two integrals

for the first one, use the substitution t =tanx

the second integral is a direct formula.




formulae on integration
 
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Friday, January 20, 2017

integral of { (cosx)^2 / [ (cosx)^2+ 4 (sinx)^2 ] }

integral of { (cos x)^2 / [ (cos x)^2+ 4 (sin x)^2 ]  }

use trigonometric formulae to change  (sin x)^2 = 1 - (cos x)^2 so that the intergral is completely in terms of (cos x)^2 .
 Now try to write the numerator in terms of the denominator.
 introduce a (-3 ) in the numerator and denominator and add and subtract 4

split it into two terms and then two integrals

The second integral contains (cos x)^2 .

divide each term with (cos x)^2 to get (sec x)^2

use trigonometric formulae tochange (sec x)^2  = 1+ (tan x)^2  in the denominator only

use substitution   t = tan x and change the limits.






formulae on integration
 
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Thursday, January 19, 2017

integral using substitution and then integration by parts

integral using substitution and then integration by parts

integral of [1 / (x^4)][sqrt( 1 + (x^2))][ log( 1 + (x^2)) - 2log( x)]

first simplify using property of logarithms

take x^2 common from the sqrt term obtain [1+1/(x^2)] and try to get the same term inside the log expression

cancel off the x to get 1 /[x^3]   then use substitution

then use integration by parts with log(t) as the first function



formulae on integration
 
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integral of sqrt{[1-sqrt(x)] / [1+sqrt(x)]}

integral of sqrt{[1-sqrt(x)] / [1+sqrt(x)]}

use the substitution sqrt(x) = cost

use trigonometric formulae to simplify [1-cos t]  and [ 1+ cos t], sint using the half angle formulae

cancel off the common factors

simplify then again use trigonometric formulae to change the square terms to first degree expressions before integrating.

again use trigonometric formulae to change the variable back to x



trigonometric identities 

formulae on integration
 
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Wednesday, January 18, 2017

integral of { [ arcsin(sqrt(x)) - arccos(sqrt(x)) ] / [ arcsin(sqrt(x)) + arccos(sqrt(x))] }

integral of { [ arcsin(sqrt(x)) - arccos(sqrt(x)) ] / [ arcsin(sqrt(x)) + arccos(sqrt(x))] }

x belongs to [0,1] 

use the result that arcsin(sqrt(x)) + arccos(sqrt(x))] = [pi / 2]
to get rid of arccos(sqrt(x)) and write the integral completely in terms of  arcsin(sqrt(x))

use a substitution to  change the arc sine function to a function involving sine function

use integration by parts to handle the new integral.






formulae on integration
 
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integral of 1 / [cos(x+a)cos(x+b)]


integral of 1 / [cos(x+a)cos(x+b)] by manipulating the angle in the numerator in terms of the angles of the denominator

introduce a  term sin(a-b) in the numerator by multiplying the numerator and denominator with it.

Then express the angle  (a-b)  in terms of the denominator(x+a) and (x+b)
using (a-b) = (x+a) - (x+b)

use trigonometric formula for expanding the numerator and then split the numerator

then simplify and integrate the two terms separately.

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Tuesday, January 17, 2017

integral of 1 / 6{ [ x^(1/2)+ x^(1/3) ] } using substitution

integral of 1 / 6{ [ x^(1/2)+ x^(1/3) ] } using substitution

here the lcm of 2 and 3 is 6


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integral of 1 / { x*sqrt[ax-x^2]} using substitution

 integral of 1 / { x*sqrt[ax-x^2]} using substitution

method is put x = (a/t) ; dx = [-a / (t^2) ] dt

replace x and dx in the given equation and simplify as shown below

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Monday, January 16, 2017

integral of (x²+1) / [(x+3)(x-1)²] using partial fractions

integral of (x²+1) / [(x+3)(x-1)²] using partial fractions



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integral using partial fractions

integral using partial fractions

integral ∫ dx / [ (x-2) (1 + x²)]



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integral of 1 / [5 + 4 cosx] with limits 0 to (pi/2)

integral of 1 / [5 + 4 cosx] with limits 0 to (pi/2)

if the denominator contains sinx  or cosx in the first degree,  the  method is to use the substitution,
t = tan(x/2) and change dx
you also have to express sin(x) and cos(x)  in terms of tan(x/2) , then use completion of squares if necessary to integrate. If there are limits it is better to change the limits




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integral ∫ { cosx/ [ sinx +cosx] } dx

integral  ∫  { cosx / [ sinx +cosx]  } dx

first express the numerator in terms of the denominator and its derivative
split into two integrals ;one  of them is of the form u' / u which will integrate to ln(u)


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