In Problems 3-8, determine whether the given function is a solution to the given differential equation.

y=sinx+x2,d2ydx2+y=x2+2

Short Answer

Expert verified

The given function is a solution to the given differential equation.

Step by step solution

01

Differentiating the given equation w.r.t. (with respect to) x

Firstly, we will differentiatey=sinx+x2 with respect to x,

dydx=cosx+2x

Again, differentiating with respect to x,

d2ydx2=-sinx+2

02

Simplification

Putting the values from step 1 in the L.H.S. (Left-hand side) of the given differential equation,

d2ydx2+y=-sinx+2+sinx+x2d2ydx2+y=x2+2

Which is the same as the R.HS. (Right-hand side) of the given differential equation.

Hence,y=sinx+x2 is a solution to the differential equationd2ydx2+y=x2+2.

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Most popular questions from this chapter

Variation of Parameters. Here is another procedure for solving linear equations that is particularly useful for higher-order linear equations. This method is called variation of parameters. It is based on the idea that just by knowing the form of the solution, we can substitute into the given equation and solve for any unknowns. Here we illustrate the method for first-order equations (see Sections 4.6 and 6.4 for the generalization to higher-order equations).

(a) Show that the general solution to (20)dydx+P(x)y=Q(x) has the formy(x)=Cyh(x)+yp(x),whereyh ( 0is a solution to equation (20) when Q(x)=0,

C is a constant, andyp(x)=v(x)yh(x) for a suitable function v(x). [Hint: Show that we can takeyh=μ-1(x) and then use equation (8).] We can in fact determine the unknown function yhby solving a separable equation. Then direct substitution of vyh in the original equation will give a simple equation that can be solved for v.

Use this procedure to find the general solution to (21) localid="1663920708127" dydx+3xy=x2, x > 0 by completing the following steps:

(b) Find a nontrivial solutionyh to the separable equation (22) localid="1663920724944" dydx+3xy=0, localid="1663920736626" x>0.

(c) Assuming (21) has a solution of the formlocalid="1663920777078" yp(x)=v(x)yh(x) , substitute this into equation (21), and simplify to obtain localid="1663920789271" v'(x)=x2yh(x).

d) Now integrate to getlocalid="1663920800433" vx

(e) Verify thatlocalid="1663920811828" y(x)=Cyh(x)+v(x)yh(x) is a general solution to (21).

Pendulum with Varying Length. A pendulum is formed by a mass m attached to the end of a wire that is attached to the ceiling. Assume that the length l(t)of the wire varies with time in some predetermined fashion. If

U(t) is the angle in radians between the pendulum and the vertical, then the motion of the pendulum is governed for small angles by the initial value problem l2(t)θ''(t)+2l(t)l'(t)θ'(t)+gl(t)sin(θ(t))=0;θ(0)=θo,θ'(0)=θ1where g is the acceleration due to gravity. Assume that l(t)=lo+l1cos(ωt-ϕ)where l1is much smaller than lo. (This might be a model for a person on a swing, where the pumping action changes the distance from the center of mass of the swing to the point where the swing is attached.) To simplify the computations, take g = 1. Using the Runge– Kutta algorithm with h = 0.1, study the motion of the pendulum when θo=0.05,θ1=0,lo=1,l1=0.1,ω=1,ϕ=0.02. In particular, does the pendulum ever attain an angle greater in absolute value than the initial angle θo?

In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

dydx=y4-x4,y(0)=7

Question: The Taylor series for f(x) =ln (x)about x2=0given in equation (13) can also be obtained as follows:

(a)Starting with the expansion 1/ (1-s) =n=0s'' and observing that

'

obtain the Taylor series for 1/xabout x0= 1.

(b)Since use the result of part (a) and termwise integration to obtain the Taylor series for f (x)=lnxaboutx0= 1.

Question: In Problems 29–34, determine the Taylor series about the point X0for the given functions and values of X0.

31. f(x)=1+x1-x.x0 = 0 ,

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