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Is the function π¦ equals three π to the π₯ minus π₯ add one a solution to the differential equation π¦ prime equals π₯ add π¦?
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Our differential equation is π¦ prime equals π₯ add π¦.
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Remember that a differential equation is an equation with a function and one or more of its derivatives.
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In this equation, we have π¦ prime.
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But remember that this is just another way of saying dπ¦ by dπ₯.
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To assess whether π¦ equals three π to the π₯ minus π₯ add one is a solution to this differential equation.
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We need to substitute it into the left-hand side of our differential equation and the right-hand side of our differential equation.
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And weβll then see whether the left-hand side agrees with the right-hand side.
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Starting with the left-hand side, if we begin with our function π¦ equals three π to the π₯ minus π₯ add one, we need to differentiate this in order to find π¦ prime.
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As we said earlier, π¦ prime is just another way of saying dπ¦ by dπ₯.
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So we need to differentiate our function π¦ with respect to π₯.
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Starting with three π to the π₯, we recall that the derivative with respect to π₯ of π to the π₯ power is π to the π₯ power.
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So three π to the π₯ differentiates to give us three π to the π₯.
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Then moving on to the next term, we know that π₯ differentiates to one.
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So negative π₯ differentiates to negative one.
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And finally, we recall that constants differentiate to zero.
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So we find that π¦ prime equals three π to the π₯ minus one.
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So thatβs the left-hand side of our differential equation when we use π¦ equals three π to the π₯ minus π₯ add one as a solution.
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Now letβs look at the right-hand side of our differential equation.
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Replacing π¦ with our function three π to the π₯ minus π₯ add one, we see that π₯ add π¦ is π₯ add three π to the π₯ minus π₯ add one.
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We then see that the π₯s cancel.
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And weβre left with three π to the π₯ add one.
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For π¦ equals three π to the π₯ minus π₯ add one to be a solution to our differential equation, we must have that the left-hand side and the right-hand side agree.
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On the left-hand side, we have three π to the π₯ minus one.
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But on the right-hand side, we have three π to the π₯ add one.
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So, in fact, we see that the left-hand side and the right-hand side of this differential equation donβt agree.
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So we can conclude that this function π¦ is not a solution to this differential equation.