Solution:
put cosecx + cotx = t =- log t = - log |cosec x +cot x | \ The solution is y= log |cosec x+ cot x| + 2x+ c 2 Giving at x = p /2 = 0, y =..
put cosecx + cotx = t =- log t = - log |cosec x +cot x | \ The solution is y= log |cosec x+ cot x| + 2x+ c 2 Giving at x = p /2 = 0, y =..Solution:
Symmetry (a) By replacing y by -y, the equation (1) is altered, therefore the curve is not symmetrical about x-axis. (b) By replacing x by -x, the equation of the curve is altered, therefore the curve is not symmetrical about y-axis. (c) Replace x and y by -x and -y respectively in the equation y ..
Solution:
f '(x) = 6x 2 - 42x + 36 f '(x) = 0 x = 1 and x = 6 are the critical values f ''(x) =12x - 42 If x =1, f ''(1) =12 - 42 = - 30 < 0 x =1 is a point of local maxima of f (x). Maximum value = 2(1) 3 - 21(1) 2 + 36(1) - 20 = -3 If x = 6, f ''(6) = 72 - 42 = 30 > 0 x = 6 is a point..
f '(x) = 6x 2 - 42x + 36 f '(x) = 0 x = 1 and x = 6 are the critical values f ''(x) =12x - 42 If x =1, f ''(1) =12 - 42 = - 30 < 0 x =1 is a point of local maxima of f (x). Maximum value = 2(1) 3 - 21(1) 2 + 36(1) - 20 = -3 If x = 6, f ''(6) = 72 - 42 = 30 > 0 x = 6 is a point..Solution:
Let us find the values of x in the given interval, for which f(x) = 0 x 3 - x 2 - 2x = 0 x(x 2 - x - 2) = 0 Partition the domain [-1, 2] to subintervals [-1, 0] and [0, 2] ..
Let us find the values of x in the given interval, for which f(x) = 0 x 3 - x 2 - 2x = 0 x(x 2 - x - 2) = 0 Partition the domain [-1, 2] to subintervals [-1, 0] and [0, 2] ..Solution:
Put t = tan - 1 x or x = tan t when x = 0 tan t = 0 when x = 1, tan t =1 2. Evaluate the definite integra..
Put t = tan - 1 x or x = tan t when x = 0 tan t = 0 when x = 1, tan t =1 2. Evaluate the definite integra..Solution:
..
..Solution:
Let m 0 be the moisture content initially and let m be the moisture content after t hours. According to problem: when t = 0, m= m 0 log m = kt + log m 0 Substituting in (2), we have \ Equation (2) becomes (i) Again when the sheet losses 95% of the moisture, m = Equation ..
Let m 0 be the moisture content initially and let m be the moisture content after t hours. According to problem: when t = 0, m= m 0 log m = kt + log m 0 Substituting in (2), we have \ Equation (2) becomes (i) Again when the sheet losses 95% of the moisture, m = Equation ..Solution:
Let the initial population be P 0 and P be the population of the colony at any instant t. Then according to the problem when t = 0, P= P 0 Equation (1) becomes log P = kt + log P 0 .(2) when P 0 is doubled, P = 2P 0 where t = 50 days From equation (2) log (2 P 0 ) = 50 k + log P 0 ..
Let the initial population be P 0 and P be the population of the colony at any instant t. Then according to the problem when t = 0, P= P 0 Equation (1) becomes log P = kt + log P 0 .(2) when P 0 is doubled, P = 2P 0 where t = 50 days From equation (2) log (2 P 0 ) = 50 k + log P 0 ..Solution:
If we replace y by , we have y'+ y \ f (x) satisfies the equation y' + y = 2 Moreover, ..
If we replace y by , we have y'+ y \ f (x) satisfies the equation y' + y = 2 Moreover, ..General solution of a differential equation
If the solution of a differential equation of order n contains n arbitrary constants, then it is called the General solution of the differential equatio..
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