Question Video: Finding the Differential Equation Using One of Its Solutions | Nagwa Question Video: Finding the Differential Equation Using One of Its Solutions | Nagwa

# Question Video: Finding the Differential Equation Using One of Its Solutions Mathematics • Higher Education

Which of the following differential equations has the solution π¦ β 2 = 5π^(π₯) + 5π^(βπ₯)? [A] π¦β² = 5π^(π₯) β 5π^(βπ₯) [B] π¦β² = 5π^(π₯) + 5π^(βπ₯) [C] π¦β² = 2π^(π₯) β 2π^(βπ₯) [D] π¦β² = 2π^(π₯) + 2π^(βπ₯) [E] π¦β² = (1/5)π^(π₯) + (1/5)π^(βπ₯)

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### Video Transcript

Which of the following differential equations has the solution π¦ minus two is equal to five π to the power of π₯ plus five π to the power of negative π₯. Option (a) π¦ prime is equal to five π to the power of π₯ minus five π to the power of negative π₯. Option (b) π¦ prime is equal to five π to the power of π₯ plus five π to the power of negative π₯. Option (c) π¦ prime is equal to two π to the power of π₯ minus two π to the power of negative π₯. Option (d) π¦ prime is equal to two π to the power of π₯ plus two π to the power of negative π₯. Or option (e) π¦ prime is equal to one-fifth π to the power of π₯ plus one-fifth π to the power of negative π₯.

The question gives us five differential equations. And it wants to know which of these differential equations has the solution π¦ minus two is equal to five π to the power of π₯ plus five π to the power of negative π₯. We recall, for this to be a solution of a differential equation, it needs to satisfy the differential equation. So this must satisfy one of the five differential equations weβre given. To start, we notice each of these five differential equations has the term π¦ prime. And π¦ is a function of π₯. So π¦ prime is equal to dπ¦ by dπ₯, the first derivative of π¦ with respect to π₯.

So because our solution needs to satisfy one of these equations, letβs find an expression for π¦ prime of the equation weβre given. So weβll start with the solution weβre given in the question. π¦ minus two is equal to five π to the power of π₯ plus five π to the power of negative π₯. We want to use this to find an expression for π¦ prime. Weβll start by rearranging this to make π¦ the subject. Weβll add two to both sides of this equation. So weβve now rewritten π¦ to be a function of π₯. This means we can find the derivative of π¦ with respect to π₯.

So to find an expression for π¦ prime, weβll differentiate with respect to π₯. This gives us π¦ prime is equal to the derivative of five π to the power of π₯ plus five π to the power of negative π₯ plus two with respect to π₯. Now we can evaluate this derivative term by term. Remember, for any constant π, the derivative of π to the power of ππ₯ with respect to π₯ is equal to π times π to the power of ππ₯.

So letβs evaluate the derivative of our first term. We can see our exponent is just π₯. So the coefficient π is equal to one. So to differentiate this term, we multiply it by one. But this doesnβt change the value. So its derivative is just five π to the power of π₯. Letβs now differentiate our second term. This time, we can see our exponent is negative π₯. This means our value of π is equal to negative one. So to differentiate this term, we need to multiply it by negative one. This gives us negative one times five π to the power of negative π₯. And of course, this just gives us negative five π to the power of negative π₯.

The last thing we need to differentiate is the constant two. However, this is a constant; it doesnβt change as our value of π₯ changes. So its rate of change is equal to zero. So weβve shown π¦ prime is equal to five π to the power of π₯ minus five π to the power of negative π₯. So letβs compare this to each of our options. First, we can see this is exactly equal to option (a). So this satisfies the differential equation in option (a). So it must be a solution to the differential equation in option (a).

We could stop here. However, for due diligence, letβs check the rest of our options. Letβs start with option (b). We can see that π¦ prime should be equal to five π to the power of π₯ plus five π to the power of negative π₯. However, we have negative five π to the power of negative π₯. So our solution does not satisfy this differential equation. So it canβt be a solution to this differential equation. So the answer cannot be option (b).

Next, we can see that option (c) tells us π¦ prime should be equal to two π to the power of π₯ minus two π to the power of negative π₯. And again, we can see this does not agree with what we have for π¦ prime of π₯. We have a coefficient of five for π to the power of π₯ and a coefficient of negative five for π to the power of negative π₯. So option (c) canβt be the correct answer. And if we check option (d) and option (e), we can see that the coefficients of π to the power of π₯ and π to the power of negative π₯ are both incorrect. This tells us that our solution cannot be a solution to either the differential equation in option (d) or the differential equation in option (e).

Therefore, we have shown only the differential equation in option (a) π¦ prime is equal to five π to the power of π₯ minus five π to the power of negative π₯ has the solution. π¦ minus two is equal to five π to the power of π₯ plus five π to the power of negative π₯.

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