Find the particular solution that satisfies the given initial condition dydx=sinx∙secy,y(π2)=π4\frac{dy}{dx}=\sin x\bullet\sec y,y\left(\frac{\pi}{2}\right)=\frac{\pi}{4}.

y = s i n − 1 ( − cos ⁡ x + 2 2 ) y=sin^{-1}\left(-\cos x+\frac{\sqrt2}{2}\right)

Find the particular solution that satisfies the given initial condition d y d x = sin x ∙ sec y , y ( π 2 ) = π 4 \frac{dy}{dx}=\sin x\bullet\sec y,y\left(\frac{\pi}{2}\right)=\frac{\pi}{4} .

we're asked to find the particular solution that satisfies the given initial condition based on the differential equation d y d x equals sine of x times secant of y, and the initial condition y of pi over two equals pi over four. So So let's go ahead and get started with our steps here to solve this separable differential equation. If D Y D X is equal to the sine of X times the secant of Y, the first thing we wanna do is separate our variables. So we wanna get our function of y and d y together on one side, our function of x and d x together on the other. So the first thing I'm gonna do here is divide over this function of y. So dividing both sides by the secant of y, remember that's the same thing as multiplying by one over the secant of y. So that will end up canceling out, and at the same time, I can go ahead and multiply both sides by d x in order to move that over as well. So now I have one over secant of y times d y equals sine of x times d x. Now one over the secant of y is a really strange way to just write cosine. So we can go ahead and just rewrite this as the cosine of y, because remember, one over the cosine of y is the secant of y. So it makes sense that one over the secant of y is the cosine of y. So we can go ahead and proceed on with our next step here, which is integrating both sides. So integrating each side here, the integral of the cosine of y dy, this is gonna give us the sine of y, and that's then equal to negative cosine of x. And then we wanna add on that constant of integration c there. So now we've successfully integrated both sides and we can move on to our next step. Here we were given an initial condition and asked to find a particular solution, so we can't skip to step four. We need to proceed with step three, which is finding c by plugging that initial condition in. Remember, the initial we are given was that y of pi over two is equal to pi over four. So that means we wanna plug pi over two in for x and pi over four in for y. So doing that here, that's the sine of pi over four is equal to negative cosine of pi over two plus c. So solving for c here, we want to evaluate these trig functions. The sine of pi over four is square root of two over two, and the cosine of pi over two is actually zero. So we get here that root two over two is equal to our constant of integration c. So now with that constant found, we can plug that back into our original equation there, our original solution. So this gives us that the sine of y is equal to negative cosine of x plus root two over two. So now that we've found c, we can go ahead and solve for y here. And we only have to do one step here, which is taking the inverse sine of each side in order to get that sine to cancel out. So on both sides, we're taking the inverse sine, then we're left with our final solution that y is equal to the inverse sine of negative cosine of x plus the square root of two over two and this gives me my particular solution to my differential equation that satisfies the given initial condition. Let us know if you have any questions and I'll see you in the next one.

Find the particular solution that satisfies the given initial condition d y d x = sin ⁡ x ∙ sec ⁡ y , y ( π 2 ) = π 4 \frac{dy}{dx}=\sin x\bullet\sec y,y\left(\frac{\pi}{2}\right)=\frac{\pi}{4} .

y = s i n − 1 ( − cos ⁡ x + 2 2 ) y=sin^{-1}\left(-\cos x+\frac{\sqrt2}{2}\right)

y = s i n − 1 ( − cos ⁡ x + π 4 ) y=sin^{-1}\left(-\cos x+\frac{\pi}{4}\right)

y = − cos ⁡ x + 2 2 y=-\cos x+\frac{\sqrt2}{2}

y = − cos ⁡ x + π 4 y=-\cos x+\frac{\pi}{4}

13. Intro to Differential Equations

Separable Differential Equations

Calculus

13. Intro to Differential Equations

Separable Differential Equations

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Multiple Choice

Find the particular solution that satisfies the given initial condition $\frac{d y}{d x} = \sin x ∙ \sec y, y (\frac{π}{2}) = \frac{π}{4}$ .

$y = s i n^{- 1} (- \cos x + \frac{\sqrt{2}}{2})$

$y = s i n^{- 1} (- \cos x + \frac{π}{4})$

$y = - \cos x + \frac{\sqrt{2}}{2}$

$y = - \cos x + \frac{π}{4}$

Verified step by step guidance

Step 1: Recognize that the given differential equation $ \frac{dy}{dx} = \sin x \cdot \sec y $ is separable. Rewrite it as $ \sec y \, dy = \sin x \ dx $. This allows us to separate the variables $ y $ and $ x $ on opposite sides of the equation.

Step 2: Integrate both sides. For the left-hand side, integrate $ \int \sec y \, dy $, which results in $ \ln|\sec y + \tan y| + C_1 $. For the right-hand side, integrate $ \int \sin x \, dx $, which results in $ -\cos x + C_2 $. Combine the constants into a single constant $ C $.

Step 3: Combine the results of the integration to form the general solution: $ \ln|\sec y + \tan y| = -\cos x + C $. Exponentiate both sides to remove the natural logarithm, yielding $ \sec y + \tan y = e^{-\cos x + C} $.

Step 4: Use the initial condition $ y(\frac{\pi}{2}) = \frac{\pi}{4} $ to solve for the constant $ C $. Substitute $ x = \frac{\pi}{2} $ and $ y = \frac{\pi}{4} $ into the equation $ \sec y + \tan y = e^{-\cos x + C} $. Simplify to find the value of $ C $.

Step 5: Substitute the value of $ C $ back into the equation and simplify to express $ y $ explicitly in terms of $ x $. Use trigonometric identities and inverse functions as needed to arrive at the particular solution.

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