| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine whether the function \( y = xe^{-x^2/2} \) satisfies the equation \( xy' = (1 - x^2)y \), we need to find the derivative of the function \( y \) with respect to \( x \) and substitute it into the differential equation. Given \( y = xe^{-x^2/2} \), we can find \( y' \) by using the product rule and the chain rule. The product rule states that \( (uv)' = u'v + uv' \), and the chain rule states that \( (f(g(x)))' = f'(g(x))g'(x) \). Let's find \( y' \): \[ y' = (xe^{-x^2/2})' \] Applying the product rule, we have: \[ y' = (x)'e^{-x^2/2} + x(e^{-x^2/2})' \] The derivative of \( x \) is \( 1 \), and to find the derivative of \( e^{-x^2/2} \), we apply the chain rule: \[ (e^{-x^2/2})' = e^{-x^2/2}(-x^2/2)' \] The derivative of \( -x^2/2 \) is \( -x \), so: \[ (e^{-x^2/2})' = e^{-x^2/2}(-x) \] Putting it all together: \[ y' = e^{-x^2/2} + x(e^{-x^2/2})(-x) \] \[ y' = e^{-x^2/2} - x^2e^{-x^2/2} \] \[ y' = e^{-x^2/2}(1 - x^2) \] Now, we substitute \( y' \) and \( y \) into the given differential equation: \[ xy' = (1 - x^2)y \] Substituting \( y' \) and \( y \): \[ x(e^{-x^2/2}(1 - x^2)) = (1 - x^2)(xe^{-x^2/2}) \] Simplifying: \[ xe^{-x^2/2}(1 - x^2) = xe^{-x^2/2}(1 - x^2) \] Since the left-hand side is equal to the right-hand side, the function \( y = xe^{-x^2/2} \) does indeed satisfy the differential equation \( xy' = (1 - x^2)y \). |
True |
| deepseekr1 | To determine if the function \( y = x e^{-x^2/2} \) satisfies the differential equation \( xy' = (1 - x^2)y \), we proceed as follows: 1. **Compute the derivative \( y' \)**: - The function \( y \) is a product of \( x \) and \( e^{-x^2/2} \). - Using the product rule: \( y' = u'v + uv' \), where \( u = x \) and \( v = e^{-x^2/2} \). - \( u' = 1 \) and \( v' = -x e^{-x^2/2} \) (using the chain rule). - Therefore, \( y' = 1 \cdot e^{-x^2/2} + x \cdot (-x e^{-x^2/2}) \). - Simplifying, \( y' = e^{-x^2/2} - x^2 e^{-x^2/2} = e^{-x^2/2}(1 - x^2) \). 2. **Substitute \( y \) and \( y' \) into the differential equation**: - The left-hand side (LHS) is \( xy' \): \[ xy' = x \cdot e^{-x^2/2}(1 - x^2) \] - The right-hand side (RHS) is \( (1 - x^2)y \): \[ (1 - x^2)y = (1 - x^2) \cdot x e^{-x^2/2} \] - Both LHS and RHS simplify to \( x(1 - x^2)e^{-x^2/2} \), confirming they are equal. 3. **Check at \( x = 0 \)**: - When \( x = 0 \), \( y = 0 \). - LHS: \( 0 \cdot y' = 0 \). - RHS: \( (1 - 0^2) \cdot 0 = 0 \). - Both sides are equal at \( x = 0 \). 4. **Verification with specific values**: - For \( x = 1 \) and \( x = 2 \), both sides of the equation yield equal results. Thus, the function \( y = x e^{-x^2/2} \) satisfies the differential equation \( xy' = (1 - x^2)y \). \[ \boxed{Yes} \] |
True |
| gpto1 | To determine whether the function \( y = x e^{-x^2/2} \) satisfies the differential equation \[ x y' = (1 - x^2) y, \] we need to compute \( y' \) and then verify if the equation holds. **Step 1: Compute \( y' \)** Given: \[ y = x e^{-x^2/2} \] We will use the product rule for differentiation, which states that if \( u(x) \) and \( v(x) \) are differentiable functions, then: \[ \frac{d}{dx}[u(x) v(x)] = u'(x) v(x) + u(x) v'(x). \] Let: - \( u(x) = x \) - \( v(x) = e^{-x^2/2} \) First, compute \( u'(x) \) and \( v'(x) \): 1. \( u'(x) = \frac{d}{dx}[x] = 1 \). 2. Compute \( v'(x) \): \[ \begin{align*} v'(x) &= \frac{d}{dx}\left[ e^{-x^2/2} \right] \\ &= e^{-x^2/2} \cdot \frac{d}{dx}\left( -\frac{x^2}{2} \right) \quad \text{(Chain Rule)} \\ &= e^{-x^2/2} \cdot \left( -x \right) \\ &= -x e^{-x^2/2}. \end{align*} \] Now, apply the product rule: \[ \begin{align*} y' &= u'(x) v(x) + u(x) v'(x) \\ &= (1) \cdot e^{-x^2/2} + x \cdot \left( -x e^{-x^2/2} \right) \\ &= e^{-x^2/2} - x^2 e^{-x^2/2}. \end{align*} \] **Step 2: Compute \( x y' \)** Multiply \( y' \) by \( x \): \[ \begin{align*} x y' &= x \left( e^{-x^2/2} - x^2 e^{-x^2/2} \right) \\ &= x e^{-x^2/2} - x^3 e^{-x^2/2} \\ &= e^{-x^2/2} x - e^{-x^2/2} x^3. \end{align*} \] **Step 3: Compute \( (1 - x^2) y \)** Given \( y = x e^{-x^2/2} \), multiply \( y \) by \( (1 - x^2) \): \[ \begin{align*} (1 - x^2) y &= (1 - x^2) \cdot x e^{-x^2/2} \\ &= x e^{-x^2/2} - x^3 e^{-x^2/2} \\ &= e^{-x^2/2} x - e^{-x^2/2} x^3. \end{align*} \] **Step 4: Compare \( x y' \) and \( (1 - x^2) y \)** From the computations above: \[ x y' = e^{-x^2/2} x - e^{-x^2/2} x^3, \] \[ (1 - x^2) y = e^{-x^2/2} x - e^{-x^2/2} x^3. \] We observe that: \[ x y' = (1 - x^2) y. \] **Conclusion:** Since \( x y' = (1 - x^2) y \), the function \( y = x e^{-x^2/2} \) satisfies the given differential equation. |
True |
| GPT-4o | To determine if the function \( y = x e^{-x^2/2} \) satisfies the equation \( xy' = (1 - x^2)y \), we need to find the derivative \( y' \) and then verify whether the given differential equation holds. First, let's find the derivative \( y' \) of the function \( y = x e^{-x^2/2} \). Using the product rule, we have: \[ y = x e^{-x^2/2} \] The product rule states: \[ (uv)' = u'v + uv' \] where \( u = x \) and \( v = e^{-x^2/2} \). First, compute \( u' \) and \( v' \): \[ u = x \quad \Rightarrow \quad u' = 1 \] \[ v = e^{-x^2/2} \quad \Rightarrow \quad v' = e^{-x^2/2} \cdot \left( -\frac{x^2}{2} \right)' = e^{-x^2/2} \cdot (-x) = -x e^{-x^2/2} \] Now apply the product rule: \[ y' = u'v + uv' \] \[ y' = (1) e^{-x^2/2} + x (-x e^{-x^2/2}) \] \[ y' = e^{-x^2/2} - x^2 e^{-x^2/2} \] \[ y' = e^{-x^2/2} (1 - x^2) \] Now we need to verify if \( xy' = (1 - x^2)y \). Substitute \( y' \) into the left-hand side of the equation: \[ xy' = x \left( e^{-x^2/2} (1 - x^2) \right) \] \[ xy' = x e^{-x^2/2} (1 - x^2) \] \[ xy' = (1 - x^2) x e^{-x^2/2} \] Compare this to the right-hand side: \[ (1 - x^2)y = (1 - x^2) \left( x e^{-x^2/2} \right) \] \[ (1 - x^2)y = (1 - x^2) x e^{-x^2/2} \] We see that: \[ xy' = (1 - x^2)y \] Therefore, the function \( y = x e^{-x^2/2} \) does indeed satisfy the differential equation \( xy' = (1 - x^2)y \). |
True |
| llama3.1-70B-Instruct | To determine if the function $y = xe^{-x^2/2}$ satisfies the equation $xy' = (1 - x^2)y$, we need to find the derivative of $y$ and then substitute both $y$ and $y'$ into the given equation. ## Step 1: Find the derivative of $y = xe^{-x^2/2}$. To find $y'$, we'll use the product rule of differentiation, which states that if $y = u \cdot v$, then $y' = u'v + uv'$. Here, $u = x$ and $v = e^{-x^2/2}$. So, $u' = 1$ and $v' = e^{-x^2/2} \cdot (-x) = -xe^{-x^2/2}$. ## Step 2: Calculate $y'$ using the product rule. $y' = u'v + uv' = 1 \cdot e^{-x^2/2} + x \cdot (-xe^{-x^2/2}) = e^{-x^2/2} - x^2e^{-x^2/2}$. ## Step 3: Substitute $y$ and $y'$ into the equation $xy' = (1 - x^2)y$. Substituting $y = xe^{-x^2/2}$ and $y' = e^{-x^2/2} - x^2e^{-x^2/2}$ into the equation gives us $x(e^{-x^2/2} - x^2e^{-x^2/2}) = (1 - x^2)xe^{-x^2/2}$. ## Step 4: Simplify both sides of the equation. Simplifying the left side gives $xe^{-x^2/2} - x^3e^{-x^2/2}$. The right side simplifies to $xe^{-x^2/2} - x^3e^{-x^2/2}$ as well. ## Step 5: Compare the simplified expressions. Both sides of the equation are identical, which means the function $y = xe^{-x^2/2}$ satisfies the equation $xy' = (1 - x^2)y$. The final answer is: $\boxed{1}$ |
True |
| llama3.1-405B-Instruct | To determine if the function $y=xe^{-x^2/2}$ satisfies the equation $xy' = (1 - x^2)y$, we need to find the derivative of $y$ with respect to $x$ and then substitute it into the equation. First, let's find the derivative of $y$: $$y' = \frac{d}{dx} (xe^{-x^2/2})$$ Using the product rule, we get: $$y' = e^{-x^2/2} + x \frac{d}{dx} (e^{-x^2/2})$$ Now, we need to find the derivative of $e^{-x^2/2}$: $$\frac{d}{dx} (e^{-x^2/2}) = -xe^{-x^2/2}$$ Substituting this back into the expression for $y'$, we get: $$y' = e^{-x^2/2} - x^2e^{-x^2/2}$$ Now, let's substitute $y'$ into the equation $xy' = (1 - x^2)y$: $$x(e^{-x^2/2} - x^2e^{-x^2/2}) = (1 - x^2)xe^{-x^2/2}$$ Expanding the left-hand side, we get: $$xe^{-x^2/2} - x^3e^{-x^2/2} = (1 - x^2)xe^{-x^2/2}$$ Simplifying, we get: $$xe^{-x^2/2} - x^3e^{-x^2/2} = xe^{-x^2/2} - x^3e^{-x^2/2}$$ This shows that the function $y=xe^{-x^2/2}$ does satisfy the equation $xy' = (1 - x^2)y$. |
True |
| DeepSeek-Coder-V2 | To determine if the function \( y = xe^{-x^2/2} \) satisfies the equation \( xy' = (1 - x^2)y \), we need to follow these steps: 1. **Find the derivative \( y' \) of the function \( y = xe^{-x^2/2} \).** Using the product rule for differentiation, we have: \[ y = x e^{-x^2/2} \] Let \( u = x \) and \( v = e^{-x^2/2} \). Then, \[ y' = u'v + uv' \] where \( u' = 1 \) and \( v' = \frac{d}{dx} e^{-x^2/2} \). To find \( v' \), we use the chain rule: \[ v' = e^{-x^2/2} \cdot \frac{d}{dx} \left( -\frac{x^2}{2} \right) = e^{-x^2/2} \cdot (-x) = -x e^{-x^2/2} \] Therefore, \[ y' = 1 \cdot e^{-x^2/2} + x \cdot (-x e^{-x^2/2}) = e^{-x^2/2} - x^2 e^{-x^2/2} \] Simplifying, we get: \[ y' = e^{-x^2/2} (1 - x^2) \] 2. **Substitute \( y' \) and \( y \) into the equation \( xy' = (1 - x^2)y \).** We have: \[ y = x e^{-x^2/2} \] and \[ y' = e^{-x^2/2} (1 - x^2) \] Now, substitute \( y' \) and \( y \) into the left-hand side of the equation: \[ xy' = x \cdot e^{-x^2/2} (1 - x^2) \] And the right-hand side of the equation is: \[ (1 - x^2)y = (1 - x^2) \cdot x e^{-x^2/2} \] Both sides are equal: \[ x \cdot e^{-x^2/2} (1 - x^2) = (1 - x^2) \cdot x e^{-x^2/2} \] Therefore, the function \( y = xe^{-x^2/2} \) satisfies the equation \( xy' = (1 - x^2)y \). The final answer is: \[ \boxed{\text{True}} \] |
True |
| Qwen2-72B-Instruct | To determine whether the function $y=xe^{-x^2/2}$ satisfies the equation $xy' = (1 - x^2)y$, we need to find the derivative of the function $y$ with respect to $x$ and substitute it into the differential equation. Given $y=xe^{-x^2/2}$, we can find $y'$ by using the product rule and the chain rule. The product rule states that $(uv)' = u'v + uv'$, and the chain rule states that $(f(g(x)))' = f'(g(x))g'(x)$. Let's find $y'$: \[ y' = (xe^{-x^2/2})' \] Applying the product rule, we have: \[ y' = (x)'e^{-x^2/2} + x(e^{-x^2/2})' \] The derivative of $x$ is $1$, and to find the derivative of $e^{-x^2/2}$, we apply the chain rule: \[ (e^{-x^2/2})' = e^{-x^2/2}(-x^2/2)' \] The derivative of $-x^2/2$ is $-x$, so: \[ (e^{-x^2/2})' = e^{-x^2/2}(-x) \] Putting it all together: \[ y' = e^{-x^2/2} + x(e^{-x^2/2})(-x) \] \[ y' = e^{-x^2/2} - x^2e^{-x^2/2} \] \[ y' = e^{-x^2/2}(1 - x^2) \] Now, we substitute $y'$ and $y$ into the given differential equation: \[ xy' = (1 - x^2)y \] Substituting $y'$ and $y$: \[ x(e^{-x^2/2}(1 - x^2)) = (1 - x^2)(xe^{-x^2/2}) \] Simplifying: \[ xe^{-x^2/2}(1 - x^2) = xe^{-x^2/2}(1 - x^2) \] Since the left-hand side is equal to the right-hand side, the function $y=xe^{-x^2/2}$ does indeed satisfy the differential equation $xy' = (1 - x^2)y$. |
True |
| baichuan4 | To determine if the function $y = xe^{-x^2/2}$ satisfies the differential equation $xy' = (1 - x^2)y$, we first need to find the derivative of $y$ with respect to $x$. The derivative of $y$ is: $$y' = \frac{d}{dx}(xe^{-x^2/2}) = e^{-x^2/2} + x \cdot \frac{d}{dx}(e^{-x^2/2})$$ $$= e^{-x^2/2} + x(-xe^{-x^2/2})$$ $$= e^{-x^2/2} - x^2e^{-x^2/2}$$ Now, let's substitute $y$ and $y'$ into the differential equation $xy' = (1 - x^2)y$: $$x(e^{-x^2/2} - x^2e^{-x^2/2}) = (1 - x^2)(xe^{-x^2/2})$$ Expanding the left-hand side (LHS) and the right-hand side (RHS) gives: $$xe^{-x^2/2} - x^3e^{-x^2/2} = xe^{-x^2/2} - x^3e^{-x^2/2}$$ We can see that both sides of the equation are identical, which means that the function $y = xe^{-x^2/2}$ does indeed satisfy the differential equation $xy' = (1 - x^2)y$. |
True |
| doubao | First find the derivative $y'$ of $y=xe^{-x^2/2}$ using the product rule and the chain rule. Let $u=x$ and $v=e^{-x^2/2}$. Then $u'=1$ and $v'=-xe^{-x^2/2}$. So $y'=u'v+uv'=1\times e^{-x^2/2}+x\times (-xe^{-x^2/2})=e^{-x^2/2}-x^2e^{-x^2/2}$. Now substitute $y=xe^{-x^2/2}$ and $y'=e^{-x^2/2}-x^2e^{-x^2/2}$ into the equation $xy'=(1-x^2)y$: $x(e^{-x^2/2}-x^2e^{-x^2/2})=(1-x^2)(xe^{-x^2/2})$. Expanding both sides gives: $xe^{-x^2/2}-x^3e^{-x^2/2}=xe^{-x^2/2}-x^3e^{-x^2/2}$, which is true. So the function $y=xe^{-x^2/2}$ satisfies the equation $xy'=(1-x^2)y$. |
True |
| minimax | To determine whether the function $y = xe^{-x^2/2}$ satisfies the differential equation $xy' = (1 - x^2)y$, we need to find the derivative of $y$ with respect to $x$ and then substitute both $y$ and $y'$ into the differential equation to check for equality. First, let's find the derivative $y'$: Given $y = xe^{-x^2/2}$, we can use the product rule to differentiate it. The product rule states that $(uv)' = u'v + uv'$, where $u = x$ and $v = e^{-x^2/2}$. The derivative of $u = x$ with respect to $x$ is $u' = 1$. The derivative of $v = e^{-x^2/2}$ with respect to $x$ can be found using the chain rule. Let's denote $w = -x^2/2$, so $v = e^w$. The derivative of $w$ with respect to $x$ is $w' = -x$. Then, the derivative of $v$ with respect to $w$ is $dv/dw = e^w$. Using the chain rule, we get: $v' = \frac{dv}{dw} \cdot \frac{dw}{dx} = e^w \cdot (-x) = -xe^{-x^2/2}$. Now, applying the product rule: $y' = u'v + uv' = (1)(e^{-x^2/2}) + (x)(-xe^{-x^2/2}) = e^{-x^2/2} - x^2e^{-x^2/2}$. Simplify $y'$: $y' = e^{-x^2/2}(1 - x^2)$. Now, let's substitute $y$ and $y'$ into the differential equation $xy' = (1 - x^2)y$: $xy' = x \cdot e^{-x^2/2}(1 - x^2)$, $(1 - x^2)y = (1 - x^2) \cdot xe^{-x^2/2}$. Both sides of the equation are equal to each other, as they are both $(1 - x^2)xe^{-x^2/2}$, so the function $y = xe^{-x^2/2}$ does indeed satisfy the differential equation $xy' = (1 - x^2)y$. |
True |
| kimi | To determine if the function \( y = xe^{-x^2/2} \) satisfies the equation \( xy' = (1 - x^2)y \), we can differentiate \( y \) with respect to \( x \) and then check if the given equation holds. Let's start by differentiating \( y \) with respect to \( x \): \[ y = xe^{-x^2/2} \] Using the product rule for differentiation, which states that \( (uv)' = u'v + uv' \), where \( u = x \) and \( v = e^{-x^2/2} \), we get: \[ y' = u'v + uv' \] \[ y' = (1)e^{-x^2/2} + x\left(-e^{-x^2/2} \cdot \frac{d}{dx}(x^2)\right) \] \[ y' = e^{-x^2/2} - x \cdot 2xe^{-x^2/2} \] \[ y' = e^{-x^2/2} - 2x^2e^{-x^2/2} \] \[ y' = e^{-x^2/2}(1 - 2x^2) \] Now, let's plug \( y \) and \( y' \) into the given equation \( xy' = (1 - x^2)y \): \[ xy' = x \cdot e^{-x^2/2}(1 - 2x^2) \] \[ (1 - x^2)y = (1 - x^2) \cdot xe^{-x^2/2} \] Now, we simplify both sides: Left-hand side: \[ x \cdot e^{-x^2/2}(1 - 2x^2) \] Right-hand side: \[ (1 - x^2) \cdot xe^{-x^2/2} \] \[ = x \cdot e^{-x^2/2}(1 - x^2) \] As we can see, both sides of the equation are equal: \[ x \cdot e^{-x^2/2}(1 - 2x^2) = x \cdot e^{-x^2/2}(1 - x^2) \] This simplifies to: \[ (1 - 2x^2) = (1 - x^2) \] However, this is not true for all \( x \). The left-hand side and the right-hand side do not match for all values of \( x \). Therefore, the function \( y = xe^{-x^2/2} \) does not satisfy the given differential equation \( xy' = (1 - x^2)y \). |
True |
| wenxinyiyan | To determine whether the function $y = xe^{-x^2/2}$ satisfies the equation $xy' = (1 - x^2)y$, we first need to find the derivative of $y$ with respect to $x$. Given function: $$ y = xe^{-x^2/2} $$ Finding the derivative $y'$: $$ y' = \frac{d}{dx}(xe^{-x^2/2}) $$ Using the product rule: $$ y' = e^{-x^2/2} + x(-x)e^{-x^2/2} $$ $$ y' = e^{-x^2/2} - x^2e^{-x^2/2} $$ Now, let's substitute $y$ and $y'$ into the given equation $xy' = (1 - x^2)y$ to check if it holds. Left side of the equation: $$ xy' = x(e^{-x^2/2} - x^2e^{-x^2/2}) $$ $$ xy' = xe^{-x^2/2} - x^3e^{-x^2/2} $$ Right side of the equation: $$ (1 - x^2)y = (1 - x^2)(xe^{-x^2/2}) $$ $$ (1 - x^2)y = xe^{-x^2/2} - x^3e^{-x^2/2} $$ Comparing both sides, we see that: $$ xy' = (1 - x^2)y $$ Therefore, the function $y = xe^{-x^2/2}$ does satisfy the equation $xy' = (1 - x^2)y$. |
True |
| zhipu | To determine whether the function \( y = xe^{-x^2/2} \) satisfies the differential equation \( xy' = (1 - x^2)y \), we need to follow these steps: 1. **Compute the derivative \( y' \) of the given function \( y \).** 2. **Substitute \( y \) and \( y' \) into the differential equation \( xy' = (1 - x^2)y \).** 3. **Simplify the expression to check if the equation holds true.** ### Step 1: Compute the derivative \( y' \) Given: \[ y = xe^{-x^2/2} \] We use the product rule for differentiation, which states: \[ (uv)' = u'v + uv' \] Let \( u = x \) and \( v = e^{-x^2/2} \). First, compute \( u' \): \[ u' = \frac{d}{dx}(x) = 1 \] Next, compute \( v' \): \[ v = e^{-x^2/2} \] \[ v' = \frac{d}{dx}(e^{-x^2/2}) \] Using the chain rule: \[ v' = e^{-x^2/2} \cdot \frac{d}{dx}(-x^2/2) \] \[ v' = e^{-x^2/2} \cdot (-x) \] \[ v' = -xe^{-x^2/2} \] Now, apply the product rule: \[ y' = u'v + uv' \] \[ y' = 1 \cdot e^{-x^2/2} + x \cdot (-xe^{-x^2/2}) \] \[ y' = e^{-x^2/2} - x^2e^{-x^2/2} \] \[ y' = (1 - x^2)e^{-x^2/2} \] ### Step 2: Substitute \( y \) and \( y' \) into the differential equation The differential equation is: \[ xy' = (1 - x^2)y \] Substitute \( y = xe^{-x^2/2} \) and \( y' = (1 - x^2)e^{-x^2/2} \): \[ x \left( (1 - x^2)e^{-x^2/2} \right) = (1 - x^2) \left( xe^{-x^2/2} \right) \] ### Step 3: Simplify the expression On the left-hand side: \[ x(1 - x^2)e^{-x^2/2} \] On the right-hand side: \[ (1 - x^2)xe^{-x^2/2} \] Notice that both sides are identical: \[ x(1 - x^2)e^{-x^2/2} = (1 - x^2)xe^{-x^2/2} \] Since the left-hand side equals the right-hand side for all \( x \), the function \( y = xe^{-x^2/2} \) indeed satisfies the differential equation \( xy' = (1 - x^2)y \). ### Conclusion The function \( y = xe^{-x^2/2} \) satisfies the equation \( xy' = (1 - x^2)y \). Therefore, the solution to the problem is: **Yes, the function \( y = xe^{-x^2/2} \) satisfies the equation \( xy' = (1 - x^2)y \).** |
True |
| gpto1 | To determine whether the function \( y = x e^{-x^2/2} \) satisfies the differential equation \[ x y' = (1 - x^2) y, \] we need to compute \( y' \) and then verify if the equation holds. **Step 1: Compute \( y' \)** Given: \[ y = x e^{-x^2/2} \] We will use the product rule for differentiation, which states that if \( u(x) \) and \( v(x) \) are differentiable functions, then: \[ \frac{d}{dx}[u(x) v(x)] = u'(x) v(x) + u(x) v'(x). \] Let: - \( u(x) = x \) - \( v(x) = e^{-x^2/2} \) First, compute \( u'(x) \) and \( v'(x) \): 1. \( u'(x) = \frac{d}{dx}[x] = 1 \). 2. Compute \( v'(x) \): \[ \begin{align*} v'(x) &= \frac{d}{dx}\left[ e^{-x^2/2} \right] \\ &= e^{-x^2/2} \cdot \frac{d}{dx}\left( -\frac{x^2}{2} \right) \quad \text{(Chain Rule)} \\ &= e^{-x^2/2} \cdot \left( -x \right) \\ &= -x e^{-x^2/2}. \end{align*} \] Now, apply the product rule: \[ \begin{align*} y' &= u'(x) v(x) + u(x) v'(x) \\ &= (1) \cdot e^{-x^2/2} + x \cdot \left( -x e^{-x^2/2} \right) \\ &= e^{-x^2/2} - x^2 e^{-x^2/2}. \end{align*} \] **Step 2: Compute \( x y' \)** Multiply \( y' \) by \( x \): \[ \begin{align*} x y' &= x \left( e^{-x^2/2} - x^2 e^{-x^2/2} \right) \\ &= x e^{-x^2/2} - x^3 e^{-x^2/2} \\ &= e^{-x^2/2} x - e^{-x^2/2} x^3. \end{align*} \] **Step 3: Compute \( (1 - x^2) y \)** Given \( y = x e^{-x^2/2} \), multiply \( y \) by \( (1 - x^2) \): \[ \begin{align*} (1 - x^2) y &= (1 - x^2) \cdot x e^{-x^2/2} \\ &= x e^{-x^2/2} - x^3 e^{-x^2/2} \\ &= e^{-x^2/2} x - e^{-x^2/2} x^3. \end{align*} \] **Step 4: Compare \( x y' \) and \( (1 - x^2) y \)** From the computations above: \[ x y' = e^{-x^2/2} x - e^{-x^2/2} x^3, \] \[ (1 - x^2) y = e^{-x^2/2} x - e^{-x^2/2} x^3. \] We observe that: \[ x y' = (1 - x^2) y. \] **Conclusion:** Since \( x y' = (1 - x^2) y \), the function \( y = x e^{-x^2/2} \) satisfies the given differential equation. |
True |
| deepseekv3 | To determine whether the function \( y = x e^{-x^2/2} \) satisfies the differential equation \( xy' = (1 - x^2)y \), we will follow these steps: 1. **Compute the derivative \( y' \) of the given function \( y \).** 2. **Substitute \( y \) and \( y' \) into the differential equation \( xy' = (1 - x^2)y \).** 3. **Simplify the equation to check if both sides are equal.** Let's go through each step in detail. ### Step 1: Compute the derivative \( y' \) Given: \[ y = x e^{-x^2/2} \] To find \( y' \), we will use the **product rule** and the **chain rule**. - **Product Rule:** If \( y = u \cdot v \), then \( y' = u' \cdot v + u \cdot v' \). - Here, \( u = x \) and \( v = e^{-x^2/2} \). Compute \( u' \) and \( v' \): - \( u' = \frac{d}{dx} (x) = 1 \) - \( v' = \frac{d}{dx} \left( e^{-x^2/2} \right) = e^{-x^2/2} \cdot \frac{d}{dx} \left( -\frac{x^2}{2} \right) = e^{-x^2/2} \cdot (-x) = -x e^{-x^2/2} \) Now, apply the product rule: \[ y' = u' \cdot v + u \cdot v' = 1 \cdot e^{-x^2/2} + x \cdot (-x e^{-x^2/2}) \] \[ y' = e^{-x^2/2} - x^2 e^{-x^2/2} \] \[ y' = e^{-x^2/2} (1 - x^2) \] ### Step 2: Substitute \( y \) and \( y' \) into the differential equation The differential equation is: \[ xy' = (1 - x^2)y \] Substitute \( y = x e^{-x^2/2} \) and \( y' = e^{-x^2/2} (1 - x^2) \): Left Side (LS): \[ xy' = x \cdot e^{-x^2/2} (1 - x^2) \] Right Side (RS): \[ (1 - x^2)y = (1 - x^2) \cdot x e^{-x^2/2} \] ### Step 3: Simplify and compare both sides Simplify the Left Side: \[ LS = x e^{-x^2/2} (1 - x^2) \] Simplify the Right Side: \[ RS = (1 - x^2) x e^{-x^2/2} \] Notice that: \[ LS = x e^{-x^2/2} (1 - x^2) \] \[ RS = x e^{-x^2/2} (1 - x^2) \] Since \( LS = RS \), the function \( y = x e^{-x^2/2} \) indeed satisfies the differential equation \( xy' = (1 - x^2)y \). ### Conclusion After computing the derivative \( y' \) and substituting into the given differential equation, we found that both sides of the equation are equal. Therefore, the function \( y = x e^{-x^2/2} \) satisfies the equation \( xy' = (1 - x^2)y \). **Final Answer:** Yes, the function \( y = x e^{-x^2/2} \) satisfies the differential equation \( xy' = (1 - x^2)y \). |
True |