m-denotes-a-fixed-nonzero-constant-and-c-is-the-constant-distinguishing-the-different-curves-in-th-3

Question

$m$ denotes a fixed nonzero constant, and $c$ is the constant distinguishing the different curves in the given family. In each case, find the equation of the orthogonal trajectories.
$$y^{2}=m x+c$$

Michael Jacobsen · Accepted Answer

for this problem were given a non zero constant M. And our goal is to find the orthogonal trajectories to this family of curves y squared equals M X plus e due to the power to on the dependent variable. Why we must differentiate implicitly with both sides on both sides with respect to X. So starting on the left hand side, the derivative of y squared turns into two y times de y dx. The driven of the right hand side will become just m itself, since C is constant and X is raised to the power one. So now if we saw for D y t X, we have that D y. T X is equal to m divide by two y. The next step in this process is now to solve the differential equation that D Y DX is equal to the negative reciprocal of that derivative we just found. So that would be equal to negative two y divide by M. So our goal now is to solve this differential equation. Let&#x27;s take this next step of dividing both sides just by why Now we obtain that one over. Why Times d y the X equals negative to over em. In our next step, we can write the following expression that the left hand side is equivalent to the derivative with respect to X of the natural log of why itself and they&#x27;ll be equal to negative two over em. The reason we can make this substitution is that the drift of the natural log brings white to the denominator. And since we&#x27;re differentiating with X respecto X implicitly, we still multiply by the D Y DX are now. Our next step is to find the anti derivative of both sides of that equation with respect to X. That gives us an immediate cancellation because the left hand side is looking for the anti derivative of a derivative. But that leaves just natural log y. So now the equation is the natural log of y is equal to negative two over em Times X plus a cost of integration. Let&#x27;s call that one upper case C. Now this log rhythmic equation can be turned into an Exponential Basie equation by writing Why equals he to the power of naked of two over m times X plus C using properties of exponents. This can also be expressed as why equals each of the power of C times each of the power of negative two over em. Times X. Let&#x27;s make one final substitution to express her answer in the form. Why equals de times each of the power of negative two over M times X, where d equals e to the power? See, then this expression will be the family of orthogonal trajectories that we were looking for.

Michael Jacobsen · Answer

in this problem were given. The M is a non zero, but fixed constant. Our goal is to find the orthogonal trajectories. The family of curves, described as y squared plus M X squared equals C, where C is a varying constant. The first step is to take the derivative of all sides of this equation implicitly, due to the fact that the dependent variable has a power of two, so free differentiate implicitly, we obtain that two y times D y DX plus two M times X will be equal to zero. Since the right hand side is constant. Recall that our goal here is to solve for D y t X. So take is our next step. Subtract two MX from both sides. Now two y de y dx equals make to em Times X, and we can divide both sides by two y. So the derivative of why is now equal to negative two mx divide by two y, and upon simplifying, we have that D Y DX is now equal to negative MX divide by why, on the next step, we&#x27;re going to be able to find those orthogonal trajectories by solving this following differential equation that D Y T X is equal to the negative of the reciprocal of the derivative we just found. So do you. Y DX, in this case, will be equal to Why divide by MX. This differential equation can be solved by first dividing both sides by y. That will give us one over. Why Times D y The X is equal to one over m Times X. Now we could make it a substitution on the left hand side to say that one over. Why Times D. Y. T X is no different than DDX of the natural log of Why Let&#x27;s give a little bit of space on this line and right that&#x27;s equal to one over m Times X. Now we conform the anti derivative of both sides of this equation with respect to X and solve. As usual, we&#x27;re looking for the anti derivative of the derivative of natural log y. So the result is just the natural lager. Their locker them off. Why itself So now the left hand side simplifies to natural log y equals by the constant rule. We have one over m, and now we&#x27;re looking for the anti drift of of one over X DX, which is known to be the natural log of X plus. The constant of integration will call Upper Casey the next step. Let&#x27;s use properties of lager of them&#x27;s allows for one over m to become a exponents on X inside the natural log function. If we do that, the natural log off. Why will be equal to the natural log of X to the power of one over M plus the constant c recalling that our goal is to sulphur. Why, if possible, Russell take as our next step converts the logarithmic equation to a base the exponential equation that gives us y equals e to the power of the natural log of X to the power of one over M plus a constant Let&#x27;s use next the power rule for exponential to show that Y is equal to eat of the power of C times E to the power of that natural log of X to the power of one over m. The next simplification we make is the fact that the Basie exponential and the natural log a rhythm function are inverse to each other. So we can simplify too. Why equals will call either the sea de and you&#x27;ll be multiplying extra power of one over M where de is equal to you to the sea. Now this family of curves will be orthogonal to the original given family.

Michael Jacobsen · Answer

for this problem were given that M is a fixed but non zero constant. A miracle is then to find the orthogonal trajectories to the curve, y equals MX plus e. To do that, we start off by taking the derivative of Why so D y DX would become M itself then the next stage of this process is now to solve the differential equation that D Y DX is the negative, reciprocal off the derivative we&#x27;ve just found, So D Y T X is negative one over m. What we need to do next is integrate both sides of this equation with respect to X. Let&#x27;s start off by re copying the equation with a little bit of space. So D Y T X is equal to negative one over m. Then integrate the left hand side with respect to X as well as the right hand side. Also with respect to X. If we look at the left hand side, it&#x27;s asking for the anti derivative of the drift of of why? What survives is just why itself, which will be equal to negative one over m times the variable X plus a constant C next. If we add negative one over M times extra bull sides of this equation. We obtain X divide by m plus. Why is equal to see and this represents the family of orthogonal trajectories to the original family of curves.

Michael Jacobsen · Answer

in this problem, we&#x27;re giving a fixed but non zero constant m. Our goal here is to find the orthogonal trajectories to this family of curb curbs. Wyche will see times X to the power of them to start off will take the derivative of why and obtain that d y dx by the power rule first. Well, right that this is equal to M times C times X to the power of M minus one through the power rule. Our next step in this process is to determine the value for that constant C since see depends on both y and X at the same time. If we divide by X to the power of them, we see that C is equal to why divide by X to the power of them and we can substitute this expression back into our derivative. So for the next step, de y, the X by substitution is equal to em times. Why divide by X to the power of them Times X to the power of and my s one. Let&#x27;s simplify this further to show that d y the X is now equal to m times. Why divide by X Our next step now is to find the family of orthogonal trajectories that could be solved by writing the differential equation. De Y. D X is equal to the negative reciprocal of the derivative that we have just found, which had become a negative X divide by and why we consult this differential equation by bringing white together and white together with D Y and X, together with DX, the differential equation that becomes m y times d y equals negative x times, DX. Then we can integrate the left hand side with respect to why and the right hand side with respect X. This produces M Divide by two y squared equals a negative X squared divide by two, plus a constant of integration. Let&#x27;s go with uppercase. See for that constant. If we like, we can add X squared divide by two to both sides of this equation. To give the slightly nicer format X squared over two plus m over two y squared equals Upper Casey. Then, if we multiply the entire equation by two, we get the result that X squared plus M y squared is equal to D, where he is equal to twice C. Then this is our family or auth organ, all trajectories to the original curve that was given

Michael Jacobsen · Answer

in this problem, we&#x27;re looking for their orthogonal trajectories to the equation. Y equals C times X squared. We start by differentiating both sides of this equation implicitly. With respect to X, the left hand side will be de y DX, and the right hand side will be too time. See Time&#x27;s X to the power of one fix expression for why Prime or D y DX involves the constant C. But see himself depends on both the function. Why, as well as an independent variable X buff we saw for C, We know that C is equal to why divide by X squared. Let&#x27;s substitute that back into our expression for the derivative D y. The X we&#x27;ll obtain that d y. The X is equal to two times of expression for C, which is why divide by X squared times X and upon simplifying this right hand side of the derivative D Y DX, well obtained to why divide by X, the next step of this process is to now solve the differential equation de y dx, which is equal to taking their original expression for the derivative and multiplying it by negative one. As well as taking the reciprocal. So we have negative X divide by two y is equal to D y DX. Now the salt this differential equation. I&#x27;m going to collect all expressions involved. Why the D. Y and the two y and all expressions involved X and bring them to opposite sides of this equation will obtain to why de y is equal to negative x DX. And if we integrate both sides of this expression, then we&#x27;ll obtain that Weiss word is equal to negative 1/2 of X squared, plus a constant of integration. Let&#x27;s call this one upper case C. The equation is essentially solved, but we can simplify it further by adding 1/2 X squared double sides. So half X squared plus y squared equals a constant C. If we&#x27;d like to simplify further, multiplying both sides of the equation by two gives us the family of curves. Ex word plus two y squared equals. Let&#x27;s use a new constant D where de is equal to two time see, and this is the format for the family of curves that&#x27;s orthogonal to the original family. Given

Michael Jacobsen · Answer

in this problem. We&#x27;re looking for the family of orthogonal trajectories to the curve given to B y squared equals two x plus e. Due to the power of to in this equation, we must differentiate both sides implicitly. If we differentiate the left hand side with respect to X, we&#x27;ll obtain to why times de y dx equals on the right hand side. We obtain simply a two since the constant C will go to zero in differentiation. We can solve this equation now for D Y dx by divine pull sides by two y so that D Y t X will be to over two. Why or de y? The X is one over Why now? Defying the family of our doggerel trajectories, we formed the differential Equation D y DX, which is equal to the negative reciprocal of the expression just found. So now D Y d X is equal to negative y. Let&#x27;s divide both sides of this differential equation by why to obtain that one over. Why Times D y. The X is equal to negative one. Then we can use the formula that the derivative with respect to X of the natural logger them of why is equal to one over Why Times de y dx. In other words, what we have over here in our differential equation can be substituted in for D by D. X of the natural log rhythm of why, although that doesn&#x27;t look like much, progress, will be able to do some cancellations on the step just after Sultry writes that D by D. X of the natural log of why is now equal to negative one. The next step is we can integrate both sides of this equation with respect to X. Let&#x27;s do that immediately at this step, and we noticed the cancellation immediately. The left hand side is looking for an anti derivative to the derivative, so all that&#x27;s left is natural log y on the left hand side, giving us a large simplification. So now natural log of why is equal to Negative X plus a constant Let&#x27;s call it upper case C. We have a log rhythmic equation. If we convert this to a base e exponential equation, we&#x27;ll obtain that Y is equal to each of the power of negative X plus C, then using properties of exponents, why is now equal to U to the power of C times each of the power of negative X Another, since substitution will allow us to write our answer in the form y equals D times each of the power of negative X, where d equals e to the C. And that&#x27;s the format for this trajectory of orthogonal curves.

Problem

$m$ denotes a fixed nonzero constant, and $c$ is …

Problem 19 Medium Difficulty

$m$ denotes a fixed nonzero constant, and $c$ is the constant distinguishing the different curves in the given family. In each case, find the equation of the orthogonal trajectories.
$$y^{2}=m x+c$$

Answer

$y=D e^{-\frac{2}{m} x}$

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$y=D e^{-\frac{2}{m} x}$

Differential Equations and Linear Algebra

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Stephen W. Goode, Scott A. AnninDifferential Equations and Linear Algebra

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