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Chapter 6: Q32E (page 327)

Given that the functionf(x)=x is a solution to y'''-x2y'+xy=0, show that the substitutiony(x)=v(x)f(x)=v(x)x reduces this equation to,xw''+3w'-x3w=0 wherew=v'.

Short Answer

Expert verified

Thus, it is proved that the given equation can be reduced toxw''+3w'-x3w=0.

Step by step solution

01

Use the given functions to reduce the given equation to xw''+3w'-x3w=0

Given thatf(x)=x is a solution toy'''-x2y'+xy=0                   ......(1)

And

y(x)=v(x)f(x)=v(x)x

Now find the derivative of y for equation (1),

y=vxy'=v+xv'

Use the valuew=v' in the above expression,

y'=v+xwy''=v'+xw'+wy''=w+xw'+wy''=2w+xw'y'''=2w'+w'+xw''y'''=3w'+xw''

02

Conclusion

Substitute the all values in the equation (1),

y'''-x2y'+xy=03w'+xw''-x2(v+xw)+x(vx)=03w'+xw''-x2v-x3w+x2v=03w'+xw''-x3w=0xw''+3w'-x3w=0

Thus, it is proved that the given equation can be reduced toxw''+3w'-x3w=0.

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Most popular questions from this chapter

Determine whether the given functions are linearly dependent or linearly independent on the interval(0,∞) .

(a){e2x,x2e2x,e-x}

(b){exsin2x,xexsin2x,ex,xex}

(c) {2e2x-ex,e2x+1,e2x-3,ex+1}

Higher-Order Cauchy–Euler Equations. A differential equation that can be expressed in the form

anxny(n)(x)+an−1xn−1y(n−1)(x)+...+a0y(x)=0

wherean,an−1,....,a0 are constants, is called a homogeneous Cauchy–Euler equation. (The second-order case is discussed in Section 4.7.) Use the substitution y=xyto help determine a fundamental solution set for the following Cauchy–Euler equations:

(a) x3y'''+x2y''−2xy'+2y=0,x>0

(b) x4y(4)+6x3y'''+2x2y''−4xy+4y=0,x>0

(c) x3y'''−2x2y''+13xy'−13y=0,x>0

[Hint: xα+βi=e(α+βi)lnx=xa{cos(βlnx)+isin(βlnx)}]

Deflection of a Beam Under Axial Force. A uniform beam under a load and subject to a constant axial force is governed by the differential equation

y(4)(x)-k2y''(x)=q(x),0<x<L,

where is the deflection of the beam, L is the length of the beam, k2is proportional to the axial force, and q(x) is proportional to the load (see Figure 6.2).

(a) Show that a general solution can be written in the form

y(x)=C1+C2x+C3ekx+C4e-kx+1k2∫q(x)xdx-xk2∫q(x)dx+ekx2k3∫q(x)e-kxdx-e-kx2k3∫q(x)ekxdx

(b) Show that the general solution in part (a) can be rewritten in the form

y(x)=c1+c2x+c3ekx+c4e-kx+∫0xq(s)G(s,x)ds,

where

G(s,x):=s-xk2-sinh[k(s-x)]k3.

(c) Let q(x)=1 First compute the general solution using the formula in part (a) and then using the formula in part (b). Compare these two general solutions with the general solution

y(x)=B1+B2x+B3ekx+B4e-kx-12k2x2,

which one would obtain using the method of undetermined coefficients.

Solve the given initial value problem

y'''−4y''+7y'−6y=0y(0)=1y'(0)=0y''(0)=0

Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem.

y'''-y''+x-1y=tanxy5=y'5=y''5=1
See all solutions

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