Mathematics, 17.07.2019 11:00 tahmidtaj150

Solve the differential equation y' =4x sqrt(1-y^2)

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Answer from: wisal96

The general solution to this differential equation is $y = \sin\left(2x^2 + c\right)$ for x such that $-\frac{\pi}{2} \le 2x^2 + c \le \frac{\pi}{2}$ ======explanation: assuming this diff. eqn. has a derivative of y with respect to x, rewrite so that we have y' = dy/dx $\dfrac{dy}{dx} = 4x \sqrt{1 - y^2}$ this is a separable differential equation as the derivative can be written as a product of two factors: a function of y and and a function x (the two variables we are differentiating). the factors are 4x and √(1-y²). to solve this differential equation for the general solution, we must separate the variables. separation of the variables can be done by treating dy/dx as a fraction. move the dx with the function-of-x factor and move the function-of-y factor with the dy: $\begin{aligned} \dfrac{dy}{dx} & = 4x \sqrt{1 - y^2} \\ \frac{1}{\sqrt{1-y^2}} \, dy & = 4x \, dx \end{aligned}$ now integrate both sides $\begin{aligned} \int \frac{1}{\sqrt{1-y^2}} \, dy & = \int 4x \, dx \end{aligned}$ on the left-hand side, note that $\frac{d}{dy} \sin^{-1}(y) = \frac{1}{\sqrt{1 - y^2}}$ so it follows that the antiderivative of $\frac{1}{\sqrt{1 - y^2}}$ is $\int \frac{1}{\sqrt{1 - y^2}}\,dy = \sin^{-1}(y) + c$ .on the right-hand side, we use the power rule for antiderivatives $\begin{aligned} \int \frac{1}{\sqrt{1-y^2}} \, dy & = \int 4x \, dx \\ \sin^{-1} (y) & = \frac{4x^2}{2} + c \\ \sin^{-1} (y) & = 2x^2 + c \end{aligned}$ note here that since by definition $-\pi/2 \le \sin^{-1} (y) \le -\pi/2$ , we must have $-\pi/2 \le 2x^2 + c \le \pi/2$ we can solve for y by taking the sine of both sides of the equation, resulting in a cancellation $\sin\left(\sin^{-1} (y) \right) = y$ on the left-hand side $\begin{aligned} \sin\left(\sin^{-1} (y) \right) & = \sin\left(2x^2 + c\right) \\ y & = \sin\left(2x^2 + c\right) \end{aligned}$ that is the general solution for $-\pi/2 \le 2x^2 + c \le \pi/2$

Answer from: Quest

c: lkllmn.

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What additional piece of information do you need to prove that lkn~knm by the sas similarity theorem

Answer from: Quest

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!