91Ó°ÊÓ

To find the first derivative, y'(x), integrate both sides of the given equation with respect to x: $$\int y^{\prime\prime}(x) dx = \int \frac{x}{(1-x^2)^{3/2}} dx$$

To integrate the right-hand side (RHS) of the equation, we can make a substitution. Let \(u = 1-x^2\), so \( du = -2x dx\). Hence, we have $$\int y^{\prime\prime}(x) dx = -\frac{1}{2}\int \frac{1}{u^{3/2}} du$$

Now, we can solve the integral on the RHS: $$-\frac{1}{2}\int \frac{1}{u^{3/2}} du = -\frac{1}{2}\int u^{-3/2} du$$ Calculate the integral: $$= -\frac{1}{2} \cdot \frac{u^{-1/2}}{-1/2} + C_1 = u^{1/2} + C_1$$ Replace \(u\) with the original expression in terms of x: $$= (1-x^2)^{1/2} + C_1$$

Now, we will integrate both sides of the equation to find the general solution, y(x): $$\int y^\prime(x) dx = \int (1-x^2)^{1/2} + C_1 dx$$

For this step, we need to integrate the function in two parts: $$y(x) = \int (1-x^2)^{1/2} dx + \int C_1 dx$$ The first integral can be solved using trigonometric substitution and the second integral is simply \(C_1x\). Let's solve the first integral: Let \(x = \sin{\theta}\), then \( dx = \cos{\theta} d\theta\). Hence, we have $$\int (1-\sin^2{\theta})^{1/2} \cos{\theta} d\theta$$ As we have \(1-\sin^2{\theta} = \cos^2{\theta}\), the integral becomes $$\int \cos^2{\theta} d\theta$$ To solve this integral, we use the double angle formula: $$\cos^2{\theta} = \frac{1 + \cos{2\theta}}{2}$$ So, the integral becomes $$\int \frac{1+\cos{2\theta}}{2} d\theta = \frac{1}{2}\int (1+\cos{2\theta}) d\theta = \frac{1}{2}(\theta + \frac{1}{2}\sin{2\theta}) + C_2$$ Convert back to x: $$\frac{1}{2}(\arcsin{x} + \frac{1}{2}x\sqrt{1-x^2}) + C_2$$

Now, we can combine the results from Step 5: $$y(x) = \frac{1}{2}(\arcsin{x} + \frac{1}{2}x\sqrt{1-x^2}) + C_2 + C_1x$$

Finally, we can rewrite the equation by combining constants \(C_2\) and \(C_1x\) into a new constant \(C\), which yields the general solution: $$y(x) = \frac{1}{2}(\arcsin{x} + \frac{1}{2}x\sqrt{1-x^2}) + C$$