Question

Докажите, что функция y= удовлетворяет соотношению 4(y')³+y"=0

Answer 1

 $y=\sqrt{\frac{x}{2}}=\frac{\sqrt{x}}{\sqrt{2}} \\ y'=\frac{1}{2\sqrt{\frac{x}{2}}}*\frac{1}{2}=\frac{1}{\sqrt{8x}}\\ y' = \frac{\sqrt{8x}}{8x}'= (8x)^{-0.5}'=-0.5*(8x)^{-1.5}*8=-\frac{1}{\sqrt{32x^3}}\\\ 4(y)'^3+y'' = 4*\frac{1}{ ( \sqrt{8x} )^3} - \frac{1}{\sqrt{32x^3}} = \\ \frac{4}{ 8\sqrt{8x^3} } - \frac{1}{2\sqrt{8x^3}} = \frac{4-4}{8\sqrt{8x^3}}=0$

Answer 2

Checking the Function for the Given Relation

To prove that the function \( y = f(x) \) satisfies the relation \( 4(y')^3 + y'' = 0 \), we need to differentiate the function and substitute the derivatives into the given relation.

Differentiating the Function

Let's start by differentiating the function \( y = f(x) \) to find \( y' \) and \( y'' \).

The first derivative of \( y = f(x) \) with respect to \( x \) is denoted as \( y' \), and the second derivative is denoted as \( y'' \).

Applying the Derivatives to the Relation

Now, let's substitute the first and second derivatives of \( y = f(x) \) into the given relation \( 4(y')^3 + y'' = 0 \) and verify if the relation holds true for the function.

I'll now proceed with the differentiation and substitution.