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

x=cos2t,dxdt+tx=sin2t

Short Answer

Expert verified

The given function is not a solution to the given differential equation.

Step by step solution

01

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

Firstly, we will differentiate x=cos2twith respect to t,

dxdt=-2sin2t

02

Simplification.

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

dxdt+tx=-2sin2t+tcos2t

which is not the same as the R.HS. (Right-hand side) of the given differential equation.

Hence,x=cos2tis not a solution to the differential equation dxdt+tx=sin2t.

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Most popular questions from this chapter

Let ϕ(x)denote the solution to the initial value problem

dydx=x-y,y(0)=1

⦁ Show that ϕ(x)=1-ϕ'(x)=1-x+ϕ(x)

⦁ Argue that the graph of ϕ is decreasing for x near zero and that as x increases from zero, ϕ(x)decreases until it crosses the line y = x, where its derivative is zero.

⦁ Let x* be the abscissa of the point where the solution curve y=ϕ(x) crosses the line y=x.Consider the sign of ϕ(x*) and argue that ϕ has a relative minimum at x*.

⦁ What can you say about the graph of y=ϕ(x) for x > x*?

⦁ Verify that y = x – 1 is a solution to dydx=x-y and explain why the graph of y=ϕ(x) always stays above the line y=x-1.

⦁ Sketch the direction field for dydx=x-y by using the method of isoclines or a computer software package.

⦁ Sketch the solution y=ϕ(x) using the direction field in part (f).

Question:(a) Use the general solution given in Example 5 to solve the IVP. 4x"+e-0.1tx=0,x(0)=1,x'(0)=-12.Also use J'0(x)=-J1(x) and Y'0(x)=-Y1(x)=-Y1(x)along withTable 6.4.1 or a CAS to evaluate coefficients.

(b) Use a CAS to graph the solution obtained in part (a) for.

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