91Ó°ÊÓ

First, let's express the second solution as \(y_2(x)=y_1(x) \cdot v(x)\), with \(y_1(x)=\sin x^2\) already given. Next, we need to determine the derivatives \(y_2^{\prime}(x)\) and \(y_2^{\prime\prime}(x)\), which we will later substitute into the given ODE. Here are the expressions for \(y_2(x)\) and its derivatives: \(y_2(x) = \sin x^2 \cdot v(x)\) \(y_2^{\prime}(x) = \cos x^2 \cdot 2x \cdot v(x) + \sin x^2 \cdot v^{\prime}(x)\) \(y_2^{\prime\prime}(x) = \cos x^2 \cdot (-4x^2 \sin x^2)\cdot v(x) + 2 [\cos x^2 \cdot 2x \cdot v(x)]^{\prime} + \sin x^2 \cdot v^{\prime\prime}(x)\)

Now, let's substitute the expressions for \(y_2(x)\), \(y_2^{\prime}(x)\), and \(y_2^{\prime\prime}(x)\) into the given ODE: \(x[\cos x^2 \cdot (-4x^2 \sin x^2)\cdot v(x) + 2 [\cos x^2 \cdot 2x \cdot v(x)]^{\prime} + \sin x^2 \cdot v^{\prime\prime}(x)]-[2x \cos x^2 \cdot v(x)+\sin x^2 \cdot v^{\prime}(x)]+4x^3\sin x^2 \cdot v(x)=0\)

In this step, we want to simplify the equation and reduce it to a first-order differential equation for \(v(x)\): \(-4x^3\sin x^2 \cos x^2 \cdot v(x) + 4x^2 \cos^2 x^2 \cdot v^{\prime}(x) + x \sin x^2 \cdot v^{\prime\prime}(x) - 2x \cos x^2 \cdot v^{\prime}(x) = 0\) Now, let's factor out \(v(x)\) and its derivatives, so we can work with an equation for \(v(x)\) only: \(v^{\prime\prime}(x) - \frac{2x \cos x^2}{x\sin x^2} v^{\prime}(x) = 0\)

Now, we solve the first-order differential equation for \(v(x)\): Let \(u(x) = v^{\prime}(x)\). Then the equation becomes: \(u^{\prime}(x) - \frac{2x \cos x^2}{x\sin x^2} u(x) = 0\) This equation is of first-order linear form, and we can use the integrating factor method to solve it. Let \(IF(x) = e^{\int \frac{-2x \cos x^2}{x\sin x^2} dx}\) be the integrating factor. This integral is hard to evaluate, but we can recognize that the differential of \(x^2\sin x^2\) is precisely the integrand: \(d(x^2\sin x^2)=-2x\cos x^2 dx\). Therefore, the integral evaluates to: \(IF(x) = e^{\ln(x^2\sin x^2)} = x^2\sin x^2\) Now, let's multiply the equation with the integrating factor: \(x^2 \sin x^2 \cdot u^{\prime}(x) - 2x \cos x^2 \cdot u(x) = 0\) The left-hand side is now the derivative of \(u(x) \cdot x^2 \sin x^2\): \( \frac{d}{dx}(u(x) \cdot x^2 \sin x^2) = 0\) Now, integrate both sides of the equation: \(u(x) \cdot x^2 \sin x^2 = C_1\) Next, remember that \(u(x)=v^{\prime}(x)\), so divide by \(x^2 \sin x^2\): \(v^{\prime}(x) = \frac{C_1}{x^2 \sin x^2}\) Integrate this equation to find \(v(x)\): \(v(x)= \int \frac{C_1}{x^2 \sin x^2} dx + C_2\) Unfortunately, this integral does not have an elementary closed form, so we must express the solution in terms of an integral: \(v(x) = \frac{C_1}{x^2 \sin x^2} dx + C_2\) Finally, we obtain the second independent solution by multiplying \(v(x)\) by \(y_1(x)\): \(y_2(x) = \sin x^2 \cdot (\int \frac{C_1}{x^2 \sin x^2} dx + C_2)\) Although the answer is expressed in an integral form, it represents the second independent solution to the given ODE.