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Show that if y(t) satisfies y"-ty=0, then y(-t) satisfies y"+ty=0.

Short Answer

Expert verified

Thus, it is proved that if y(t)satisfies y"-ty=0, theny(−t) satisfiesy"+ty=0.

Step by step solution

01

General form

Chain rule of the derivative states that:

ddx[f(g(x))]=f'(g(x))×g'(x)

02

Evaluate the given equation.

Given thaty(t)satisfies y"-ty=0.

To prove that y(−t)satisfies y"+ty=0.

Let us assume that, u(t)=y(-t).

Using the chain rulefind the first and second derivatives of u(t).

Case (1):

u'(t)=ddt[y(-t)]=y'(-t)×(-t)'=-y'(-t)

Case (2):

u"(t)=[-y'(-t)]'=-y"(-t)×(-t)'=y"(-t)

03

Substitute the values.

Substitute t within y"+ty=0.

y"(-t)-(-t)y(-t)=0y"(-t)+ty(-t)=0

Substitute the result of cases (1) and (2);

u"(t)+tu(t)=0

Therefore,u(t)=y(-t) satisfies the equation y"+ty=0.

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