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

20%Step-by-step explanation: Basically you find the difference which is twelve, and put it over the original amount which is 60. You get 1/5 which in percent form is 20%...

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

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noslengerman

17.07.2019

Mathematics

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 $-\pi/2 \le 2x^2 + c \le \pi/2$

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

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