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Determine the function π of π‘ such that π prime of π‘ is equal to negative two divided by three times π‘ squared plus one and π evaluated at one is equal to zero.
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The question wants us to determine the function π of π‘.
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The information weβre given is that the derivative of π of π‘ with respect to π‘ is equal to negative two divided by three times π‘ squared plus one and π evaluated at one is equal to zero.
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This is a differential equation.
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In fact, itβs a simple differential equation because weβre just given π prime of π‘ in terms of π‘.
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So to solve this, weβre going to integrate both sides of this equation with respect to π‘.
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This will give us our function π of π‘ up to a constant of integration.
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This gives us the integral of π prime of π‘ with respect to π‘ is equal to the integral of negative two divided by three times π‘ squared plus one with respect to π‘.
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Remember, integration and differentiation are opposite processes.
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So, the integral of π prime of π‘ with respect to π‘ is π of π‘ up to a constant of integration.
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So to find π of π‘, we just need to evaluate our integral.
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Before we evaluate our integral, weβll take out our constant factor of negative two divided by three.
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This gives us negative two over three times the integral of one divided by π‘ squared plus one with respect to π‘.
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But this integral is just one of our standard integrals involving inverse trigonometric functions.
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We know the integral of one divided by π₯ squared plus one with respect to π₯ is equal to the inverse tan of π₯ plus the constant of integration πΆ.
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So, we can just apply this to evaluate our integral.
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We get negative two over three times the inverse tan of π‘ plus πΆ.
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So, this is the general solution to our differential equation.
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Remember, we want the solution where π evaluated at one is equal to zero.
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So, weβll substitute this information into our general solution to find the value of πΆ.
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This gives us zero is equal to negative two-thirds times the inverse tan of one plus πΆ.
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We can simplify this equation.
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We can divide both sides through by negative two-thirds.
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Next, we can evaluate the inverse tan of one; itβs equal to π by four.
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Finally, we subtract π by four from both sides of this equation.
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We see that πΆ is equal to negative π by four.
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All we need to do now is use this value of πΆ in our general solution.
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Doing this, we get π of π‘ is equal to negative two over three times the inverse tan of π‘ minus π over four.
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And we could leave our answer like this.
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However, weβll distribute negative two-thirds over our parentheses.
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This gives us negative two times the inverse tan of π‘ divided by three plus π by six.
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And this gives us our final answer.
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Therefore, weβve shown if π prime of π‘ is equal to negative two over three times π‘ squared plus one and π of one is equal to zero.
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Then π of π‘ is equal to negative two times the inverse tan of π‘ divided by three minus π by six.