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Sagot :
Certainly! Let us start by defining the function [tex]\( y = \sin^{-1}(x) \)[/tex]. To show that the given differential equation holds, we will compute the first and second derivatives of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] and then substitute them into the differential equation.
### Step 1: First Derivative of [tex]\( y \)[/tex]
Given:
[tex]\[ y = \sin^{-1}(x) \][/tex]
We find the first derivative [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ \frac{d}{dx} \left[\sin^{-1}(x)\right] = \frac{1}{\sqrt{1 - x^2}} \][/tex]
### Step 2: Second Derivative of [tex]\( y \)[/tex]
Next, we find the derivative of the first derivative [tex]\(\frac{dy}{dx}\)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} \left(\frac{1}{\sqrt{1 - x^2}}\right) \][/tex]
We use the chain rule to differentiate this expression. Let:
[tex]\[ u = 1 - x^2 \][/tex]
Then:
[tex]\[ \frac{d}{dx} \left(\frac{1}{\sqrt{u}}\right) = \frac{d}{du} \left(u^{-\frac{1}{2}}\right) \cdot \frac{du}{dx} \][/tex]
First compute [tex]\(\frac{d}{du} \left(u^{-\frac{1}{2}}\right)\)[/tex]:
[tex]\[ \frac{d}{du} \left(u^{-\frac{1}{2}}\right) = -\frac{1}{2} u^{-\frac{3}{2}} \][/tex]
Next, compute [tex]\(\frac{du}{dx}\)[/tex]:
[tex]\[ \frac{du}{dx} = \frac{d}{dx} \left(1 - x^2\right) = -2x \][/tex]
Combining these results:
[tex]\[ \frac{d}{dx} \left(\frac{1}{\sqrt{1 - x^2}}\right) = -\frac{1}{2}(1 - x^2)^{-\frac{3}{2}} \cdot (-2x) = \frac{x}{(1 - x^2)^{\frac{3}{2}}} \][/tex]
Thus, the second derivative is:
[tex]\[ \frac{d^2 y}{dx^2} = \frac{x}{(1 - x^2)^{\frac{3}{2}}} \][/tex]
### Step 3: Substitute into the Given Expression
Now we substitute [tex]\(\frac{dy}{dx}\)[/tex] and [tex]\(\frac{d^2 y}{dx^2}\)[/tex] into the given differential equation:
[tex]\[ \left(1 - x^2\right) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0 \][/tex]
Substituting the values, we get:
[tex]\[ \left(1 - x^2\right) \left(\frac{x}{(1 - x^2)^{\frac{3}{2}}}\right) - x \left(\frac{1}{\sqrt{1 - x^2}}\right) \][/tex]
Simplify each term:
1. For the first term:
[tex]\[ \left(1 - x^2\right) \left(\frac{x}{(1 - x^2)^{\frac{3}{2}}}\right) = \frac{x (1 - x^2)}{(1 - x^2)^{\frac{3}{2}}} = \frac{x}{(1 - x^2)^{\frac{1}{2}}} \][/tex]
2. For the second term:
[tex]\[ x \left(\frac{1}{\sqrt{1 - x^2}}\right) = \frac{x}{(1 - x^2)^{\frac{1}{2}}} \][/tex]
We see that both terms are the same:
[tex]\[ \frac{x}{(1 - x^2)^{\frac{1}{2}}} - \frac{x}{(1 - x^2)^{\frac{1}{2}}} = 0 \][/tex]
Thus:
[tex]\[ 0 = 0 \][/tex]
This completes the proof that:
[tex]\[ \left(1 - x^2\right) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0 \][/tex]
Hence, the given differential equation holds true for [tex]\( y = \sin^{-1}(x) \)[/tex].
### Step 1: First Derivative of [tex]\( y \)[/tex]
Given:
[tex]\[ y = \sin^{-1}(x) \][/tex]
We find the first derivative [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ \frac{d}{dx} \left[\sin^{-1}(x)\right] = \frac{1}{\sqrt{1 - x^2}} \][/tex]
### Step 2: Second Derivative of [tex]\( y \)[/tex]
Next, we find the derivative of the first derivative [tex]\(\frac{dy}{dx}\)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} \left(\frac{1}{\sqrt{1 - x^2}}\right) \][/tex]
We use the chain rule to differentiate this expression. Let:
[tex]\[ u = 1 - x^2 \][/tex]
Then:
[tex]\[ \frac{d}{dx} \left(\frac{1}{\sqrt{u}}\right) = \frac{d}{du} \left(u^{-\frac{1}{2}}\right) \cdot \frac{du}{dx} \][/tex]
First compute [tex]\(\frac{d}{du} \left(u^{-\frac{1}{2}}\right)\)[/tex]:
[tex]\[ \frac{d}{du} \left(u^{-\frac{1}{2}}\right) = -\frac{1}{2} u^{-\frac{3}{2}} \][/tex]
Next, compute [tex]\(\frac{du}{dx}\)[/tex]:
[tex]\[ \frac{du}{dx} = \frac{d}{dx} \left(1 - x^2\right) = -2x \][/tex]
Combining these results:
[tex]\[ \frac{d}{dx} \left(\frac{1}{\sqrt{1 - x^2}}\right) = -\frac{1}{2}(1 - x^2)^{-\frac{3}{2}} \cdot (-2x) = \frac{x}{(1 - x^2)^{\frac{3}{2}}} \][/tex]
Thus, the second derivative is:
[tex]\[ \frac{d^2 y}{dx^2} = \frac{x}{(1 - x^2)^{\frac{3}{2}}} \][/tex]
### Step 3: Substitute into the Given Expression
Now we substitute [tex]\(\frac{dy}{dx}\)[/tex] and [tex]\(\frac{d^2 y}{dx^2}\)[/tex] into the given differential equation:
[tex]\[ \left(1 - x^2\right) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0 \][/tex]
Substituting the values, we get:
[tex]\[ \left(1 - x^2\right) \left(\frac{x}{(1 - x^2)^{\frac{3}{2}}}\right) - x \left(\frac{1}{\sqrt{1 - x^2}}\right) \][/tex]
Simplify each term:
1. For the first term:
[tex]\[ \left(1 - x^2\right) \left(\frac{x}{(1 - x^2)^{\frac{3}{2}}}\right) = \frac{x (1 - x^2)}{(1 - x^2)^{\frac{3}{2}}} = \frac{x}{(1 - x^2)^{\frac{1}{2}}} \][/tex]
2. For the second term:
[tex]\[ x \left(\frac{1}{\sqrt{1 - x^2}}\right) = \frac{x}{(1 - x^2)^{\frac{1}{2}}} \][/tex]
We see that both terms are the same:
[tex]\[ \frac{x}{(1 - x^2)^{\frac{1}{2}}} - \frac{x}{(1 - x^2)^{\frac{1}{2}}} = 0 \][/tex]
Thus:
[tex]\[ 0 = 0 \][/tex]
This completes the proof that:
[tex]\[ \left(1 - x^2\right) \frac{d^2 y}{dx^2} - x \frac{dy}{dx} = 0 \][/tex]
Hence, the given differential equation holds true for [tex]\( y = \sin^{-1}(x) \)[/tex].
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