Solve the differential equation by variation of parameters. y'' + y = csc x Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on.

Solve the given differential equation by VARIATION of PARAMETERS. y''+y=cscx Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on.

2017年12月20日 · Domain of y=csc(x) is x\\inRR, x\\ne pi*n, n\\inZZ. Range of y=csc(x) is y<=-1 or y>=1. y=csc(x) is the reciprocal of y=sin(x) so its domain and range are related to sine's domain and range. Since the range of y=sin(x) is -1<=y<=1 we get that the range of y=csc(x) is y<=-1 or y>=1, which encompasses the reciprocal of every value in the range of sine. The domain of y=csc(x) is every value in ...

Question 1167385: Solve the differential equation: y’ = csc x - y cot x * Answer by solver91311(24713) ( Show Source ): You can put this solution on YOUR website!

Question 6 Find the derivative. y = (csc x + cot x)(CSC X - cot x) O y'= - CSC X cot x O y'= 1 O y'= 0 O y'= - CSC2 x Question 7 Find the derivative of the function. q = V12r - 15 1 2012r - 15 o 12-514 2V12r - 15 -5r4 V12r-15 1 2V12-5r4 Question 8 Find dy/dt. y = t2(t + 225 O 9t®(+4 + 2)4(20+4 + 2) Otºtº + 2)^(2913 +18) 180t34(+4 + 2)4 O t8(+4 + 2)4(29+4 + 18) Question 9 Find the derivative ...

The derivatives of \sec(x), \cot(x), and \csc(x) can be calculated by using the quotient rule of differentiation together with the identities \sec(x)=\frac{1}{\cos(x ...

2014年9月27日 · The range of cscx is (-infty,-1]cup[1,infty). Let us look at some details. We know |sinx| le 1 by taking the reciprocal, Rightarrow |cscx|=1/|sinx| ge 1 (Note that we need to switch the direction of inequality.)

2014年8月30日 · By Product Rule, we can find: y'=-csc(x)cot^2(x)-csc^3(x) Remember: [scs(x)]'=-csc(x)cot(x) [cot(x)]'=-csc^2(x) By Product Rule, y'=[-csc(x)cot(x)]cdot cot(x)+csc(x ...

2014年9月4日 · csc(x) =1/sin(x) Vertical asymptotes happen where the equation gives an undefined answer is when x/0 where x in RR and x !=0. There for, when sin(x) = 0 then there is a vertical asymptote.

Question: Consider the following differential equation to be solved by variation of parameters. y + y = csc(x) Find the complementary function of the differential equation. Find the general solution of the differential equation. Y(x) = Consider the following differential equation to be solved by variation of parameters.