| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To determine if the given differential equation is the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$, we need to compute the total derivative of $\phi$ and compare it to the given differential equation. The total derivative of $\phi$ with respect to $t$ is given by: $$ \frac{d\phi}{dt} = \frac{\partial\phi}{\partial t} + \frac{\partial\phi}{\partial y}\frac{dy}{dt} $$ First, we compute the partial derivatives of $\phi$ with respect to $t$ and $y$: $$ \frac{\partial\phi}{\partial t} = 2t + y^2 $$ $$ \frac{\partial\phi}{\partial y} = 2ty $$ Next, we substitute these partial derivatives into the expression for the total derivative: $$ \frac{d\phi}{dt} = (2t + y^2) + (2ty)\frac{dy}{dt} $$ Now, we simplify the expression for the total derivative: $$ \frac{d\phi}{dt} = 2t + y^2 + 2ty\frac{dy}{dt} $$ Comparing this to the given differential equation, we see that they are not the same: $$ 2ty\frac{dy}{dt} + 2t + ty^2 = 0 $$ $$ 2t + y^2 + 2ty\frac{dy}{dt} \neq 2ty\frac{dy}{dt} + 2t + ty^2 $$ Therefore, the given differential equation is not the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$. However, we can see that the differential equation is closely related to the total derivative. In fact, the given differential equation can be written as: $$ 2ty\frac{dy}{dt} + 2t + ty^2 = 0 $$ $$ \Rightarrow 2ty\frac{dy}{dt} = -(2t + ty^2) $$ $$ \Rightarrow 2ty\frac{dy}{dt} = -\frac{\partial\phi}{\partial t} $$ So, the given differential equation is the negative of the partial derivative of $\phi$ with respect to $t$ multiplied by $2y$. This suggests that the given differential equation may be related to a conservative vector field, but it is not the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$. |
False |
| deepseekr1 | To determine if the differential equation \(2tyy' + 2t + ty^2 = 0\) is the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\), we need to compute the total derivative of \(\phi\) and compare it to the given differential equation. First, we compute the partial derivatives of \(\phi\): - The partial derivative of \(\phi\) with respect to \(t\) is \(\frac{\partial \phi}{\partial t} = 2t + y^2\). - The partial derivative of \(\phi\) with respect to \(y\) is \(\frac{\partial \phi}{\partial y} = 2ty\). Using these partial derivatives, the total derivative of \(\phi\) with respect to \(t\) is: \[ \frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial y} \cdot \frac{dy}{dt} = (2t + y^2) + 2ty \cdot y' \] Simplifying this, we get: \[ \frac{d\phi}{dt} = 2t + y^2 + 2tyy' \] The given differential equation is: \[ 2tyy' + 2t + ty^2 = 0 \] Comparing the total derivative \(\frac{d\phi}{dt} = 2t + y^2 + 2tyy'\) with the given differential equation, we notice that the term involving \(y^2\) differs. In the total derivative, the term is \(y^2\), whereas in the differential equation, it is \(ty^2\). This discrepancy means the given differential equation is not the total derivative of \(\phi(t, y)\). Thus, the final answer is: \[ \boxed{\text{No}} \] |
False |
| gpto1 | To determine whether the given differential equation is the total derivative of the potential function \(\phi(t, y) = t^2 + t y^2\), we'll compute the total derivative \(\frac{d\phi}{dt}\) along the solutions \(y(t)\) and compare it to the left-hand side of the differential equation. If they match, then the differential equation represents the total derivative of \(\phi\). Let's start by computing the total derivative \(\frac{d\phi}{dt}\): 1. **Compute the partial derivatives of \(\phi\):** \[ \frac{\partial \phi}{\partial t} = 2t + y^2, \quad \frac{\partial \phi}{\partial y} = 2 t y \] 2. **Compute the total derivative:** \[ \frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial y} \cdot y' = \left(2t + y^2\right) + \left(2 t y\right) y' = 2 t y y' + 2 t + y^2 \] Now, let's write the given differential equation: \[ 2 t y y' + 2 t + t y^2 = 0 \] Notice that in the expression for \(\frac{d\phi}{dt}\), the last term is \(y^2\), whereas in the differential equation, the last term is \(t y^2\). This discrepancy shows that the left-hand side of the differential equation is not exactly equal to \(\frac{d\phi}{dt}\). **Comparison:** \[ \frac{d\phi}{dt} = 2 t y y' + 2 t + y^2 \quad \text{(from total derivative)} \] \[ 2 t y y' + 2 t + t y^2 \quad \text{(from differential equation)} \] Subtracting the two expressions, we find: \[ \left(2 t y y' + 2 t + y^2\right) - \left(2 t y y' + 2 t + t y^2\right) = y^2 - t y^2 = y^2 (1 - t) \] Since this difference is not zero (unless \(t = 1\), which is not generally the case), the given differential equation does not represent the total derivative of \(\phi\). **Conclusion:** No, because when we compute the total derivative of \(\phi(t, y)\), it does not match the equation; thus, the differential equation is not the total derivative of \(\phi\). |
False |
| GPT-4o | To determine if the differential equation \(2tyy' + 2t + ty^2 = 0\) is the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\), we need to compute the total derivative of \(\phi(t, y)\) with respect to \(t\) and compare it to the given differential equation. First, let's find the total derivative of \(\phi(t, y)\) with respect to \(t\): \[ \phi(t, y) = t^2 + ty^2 \] The total derivative of \(\phi\) with respect to \(t\) is given by: \[ \frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial y} \cdot \frac{dy}{dt} \] We need to compute the partial derivatives \(\frac{\partial \phi}{\partial t}\) and \(\frac{\partial \phi}{\partial y}\): \[ \frac{\partial \phi}{\partial t} = \frac{\partial}{\partial t} (t^2 + ty^2) = 2t + y^2 \] \[ \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} (t^2 + ty^2) = 2ty \] Now, substituting these partial derivatives into the formula for the total derivative, we get: \[ \frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial y} \cdot \frac{dy}{dt} = (2t + y^2) + (2ty) \cdot y' \] Rewriting this, we have: \[ \frac{d\phi}{dt} = 2t + y^2 + 2tyy' \] The given differential equation is: \[ 2tyy' + 2t + ty^2 = 0 \] Let's compare the total derivative \(\frac{d\phi}{dt}\) with the given differential equation: \[ \frac{d\phi}{dt} = 2t + y^2 + 2tyy' \] \[ 2tyy' + 2t + ty^2 = 0 \] We can see that the terms \(2t\) and \(2tyy'\) match, but the term \(ty^2\) in the differential equation is not in the total derivative as is. Instead, the total derivative has \(y^2\) without the \(t\) coefficient. It appears there is a discrepancy in the terms \(y^2\) and \(ty^2\). To resolve this discrepancy, note that the given differential equation can be rewritten by factoring out \(t\): \[ 2tyy' + 2t + ty^2 = t(2yy' + 2 + y^2) = 0 \] For the differential equation to hold, either \(t = 0\) or \(2yy' + 2 + y^2 = 0\). We are interested in the case when \(t \neq 0\): \[ 2yy' + 2 + y^2 = 0 \implies 2tyy' + 2t + ty^2 = 0 \] Thus, the differential equation is equivalent to: \[ \frac{d\phi}{dt} = 0 \] This implies that \(\phi(t, y)\) is constant along the solutions of the differential equation. Therefore, the differential equation \(2tyy' + 2t + ty^2 = 0\) is indeed the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\). |
False |
| llama3.1-70B-Instruct | To determine if the differential equation $2tyy' + 2t + ty^2 = 0$ is the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$, we need to find the total derivative of $\phi(t, y)$ and compare it with the given differential equation. ## Step 1: Find the partial derivatives of $\phi(t, y)$ with respect to $t$ and $y$. The partial derivative of $\phi(t, y)$ with respect to $t$ is $\frac{\partial \phi}{\partial t} = 2t + y^2$. The partial derivative of $\phi(t, y)$ with respect to $y$ is $\frac{\partial \phi}{\partial y} = 2ty$. ## Step 2: Find the total derivative of $\phi(t, y)$. The total derivative of $\phi(t, y)$ is given by $\frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial y} \frac{dy}{dt} = (2t + y^2) + (2ty) \frac{dy}{dt}$. ## Step 3: Compare the total derivative with the given differential equation. The given differential equation is $2tyy' + 2t + ty^2 = 0$. We can rewrite this as $2ty \frac{dy}{dt} + 2t + ty^2 = 0$. Comparing this with the total derivative, we see that they are the same. The final answer is: $\boxed{0}$ |
False |
| llama3.1-405B-Instruct | To determine if the differential equation $2tyy' + 2t + ty^2 = 0$ is the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$, we need to find the total derivative of $\phi(t, y)$ and compare it with the given differential equation. The total derivative of $\phi(t, y)$ is given by: $$\frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial y} \frac{dy}{dt}$$ where $\frac{\partial \phi}{\partial t}$ and $\frac{\partial \phi}{\partial y}$ are the partial derivatives of $\phi(t, y)$ with respect to $t$ and $y$, respectively. First, let's find the partial derivatives of $\phi(t, y)$: $$\frac{\partial \phi}{\partial t} = 2t + y^2$$ $$\frac{\partial \phi}{\partial y} = 2ty$$ Now, we can substitute these partial derivatives into the formula for the total derivative: $$\frac{d\phi}{dt} = (2t + y^2) + (2ty) \frac{dy}{dt}$$ Simplifying the expression, we get: $$\frac{d\phi}{dt} = 2t + y^2 + 2tyy'$$ Comparing this expression with the given differential equation $2tyy' + 2t + ty^2 = 0$, we can see that they are not identical. The differential equation has an additional term $ty^2$ that is not present in the total derivative of $\phi(t, y)$. Therefore, the differential equation $2tyy' + 2t + ty^2 = 0$ is not the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$. |
False |
| DeepSeek-Coder-V2 | To determine if the differential equation \(2tyy' + 2t + ty^2 = 0\) is the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\), we need to follow these steps: 1. **Compute the partial derivatives of the potential function \(\phi(t, y)\):** \[ \phi(t, y) = t^2 + ty^2 \] - Partial derivative with respect to \(t\): \[ \frac{\partial \phi}{\partial t} = 2t + y^2 \] - Partial derivative with respect to \(y\): \[ \frac{\partial \phi}{\partial y} = 2ty \] 2. **Form the total derivative \(d\phi\) using the chain rule:** The total derivative \(d\phi\) is given by: \[ d\phi = \frac{\partial \phi}{\partial t} dt + \frac{\partial \phi}{\partial y} dy \] Substituting the partial derivatives we found: \[ d\phi = (2t + y^2) dt + (2ty) dy \] 3. **Compare the given differential equation with the total derivative:** The given differential equation is: \[ 2tyy' + 2t + ty^2 = 0 \] Rewrite \(y'\) as \(\frac{dy}{dt}\): \[ 2ty \frac{dy}{dt} + 2t + ty^2 = 0 \] Rearrange to match the form of the total derivative: \[ 2ty \frac{dy}{dt} = -2t - ty^2 \] Multiply both sides by \(dt\): \[ 2ty dy = (-2t - ty^2) dt \] This can be rewritten as: \[ (2t + y^2) dt + 2ty dy = 0 \] 4. **Conclusion:** The rearranged differential equation \((2t + y^2) dt + 2ty dy = 0\) matches the form of the total derivative \(d\phi = (2t + y^2) dt + (2ty) dy\). Therefore, the differential equation \(2tyy' + 2t + ty^2 = 0\) is indeed the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\). \[ \boxed{\text{True}} \] |
False |
| Qwen2-72B-Instruct | To determine if the given differential equation is the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$, we need to compute the total derivative of $\phi$ and compare it to the given differential equation. The total derivative of $\phi$ with respect to $t$ is given by: $$ \frac{d\phi}{dt} = \frac{\partial\phi}{\partial t} + \frac{\partial\phi}{\partial y}\frac{dy}{dt} $$ First, we compute the partial derivatives of $\phi$ with respect to $t$ and $y$: $$ \frac{\partial\phi}{\partial t} = 2t + y^2 $$ $$ \frac{\partial\phi}{\partial y} = 2ty $$ Next, we substitute these partial derivatives into the expression for the total derivative: $$ \frac{d\phi}{dt} = (2t + y^2) + (2ty)\frac{dy}{dt} $$ Now, we simplify the expression for the total derivative: $$ \frac{d\phi}{dt} = 2t + y^2 + 2ty\frac{dy}{dt} $$ Comparing this to the given differential equation, we see that they are not the same: $$ 2ty\frac{dy}{dt} + 2t + ty^2 = 0 $$ $$ 2t + y^2 + 2ty\frac{dy}{dt} \neq 2ty\frac{dy}{dt} + 2t + ty^2 $$ Therefore, the given differential equation is not the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$. However, we can see that the differential equation is closely related to the total derivative. In fact, the given differential equation can be written as: $$ 2ty\frac{dy}{dt} + 2t + ty^2 = 0 $$ $$ \Rightarrow 2ty\frac{dy}{dt} = -(2t + ty^2) $$ $$ \Rightarrow 2ty\frac{dy}{dt} = -\frac{\partial\phi}{\partial t} $$ So, the given differential equation is the negative of the partial derivative of $\phi$ with respect to $t$ multiplied by $2y$. This suggests that the given differential equation may be related to a conservative vector field, but it is not the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$. |
False |
| baichuan4 | To determine if the differential equation $2tyy' + 2t + ty^2 = 0$ is the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$, we need to calculate the total derivative of $\phi(t, y)$ with respect to $t$. The total derivative of $\phi(t, y)$ with respect to $t$ is given by: $$ \frac{d\phi}{dt} = \frac{\partial\phi}{\partial t} + \frac{\partial\phi}{\partial y}\frac{dy}{dt} $$ Calculating the partial derivatives of $\phi(t, y)$: $$ \frac{\partial\phi}{\partial t} = 2t + y^2 $$ $$ \frac{\partial\phi}{\partial y} = 2ty $$ Substituting these into the total derivative equation: $$ \frac{d\phi}{dt} = (2t + y^2) + (2ty)\frac{dy}{dt} $$ Rearranging to match the form of the given differential equation: $$ 2ty\frac{dy}{dt} + 2t + y^2 = 0 $$ Since this matches the given differential equation $2tyy' + 2t + ty^2 = 0$ (with $y'$ representing $\frac{dy}{dt}$), we can conclude that the differential equation is indeed the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$. |
False |
| doubao | First find the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$. $\frac{\partial \phi}{\partial t} = 2t + y^2$ and $\frac{\partial \phi}{\partial y} = 2ty$. The total derivative is $d\phi = (2t + y^2)dt + 2ty dy$. Now let's look at the given differential equation $2tyy' + 2t + ty^2 = 0$. Rewrite it as $2t(y\frac{dy}{dt}) + 2t + ty^2 = 0$. This is not equal to the total derivative we found. So the answer is no. |
False |
| minimax | To determine if the given differential equation is the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$, we need to compute the partial derivatives of $\phi$ with respect to $t$ and $y$ and then see if they correspond to the terms in the differential equation. The potential function $\phi(t, y) = t^2 + ty^2$ has the following partial derivatives: 1. The partial derivative of $\phi$ with respect to $t$ is: $$ \frac{\partial \phi}{\partial t} = 2t + y^2 $$ 2. The partial derivative of $\phi$ with respect to $y$ is: $$ \frac{\partial \phi}{\partial y} = 2ty $$ Now, the differential equation given is $2tyy' + 2t + ty^2 = 0$. To see if this is the total derivative of $\phi$, we need to express it in terms of the partial derivatives of $\phi$. The total derivative of $\phi$ with respect to $t$ should be: $$ d\phi = \frac{\partial \phi}{\partial t} dt + \frac{\partial \phi}{\partial y} dy $$ Substituting the partial derivatives we found into this expression gives us: $$ d\phi = (2t + y^2) dt + (2ty) dy $$ Now, we can rewrite this expression to make it look like the differential equation. To do this, we need to divide by $dt$ and rearrange terms: $$ \frac{d\phi}{dt} = 2t + y^2 + 2ty \frac{dy}{dt} $$ Comparing this with the given differential equation, we see that the terms do not match. The term $2ty \frac{dy}{dt}$ corresponds to $2tyy'$ in the differential equation, but the constant term $2t$ does not have a corresponding term in the differential equation. Therefore, the given differential equation is not the total derivative of the potential function $\phi(t, y) = t^2 + ty^2$. |
False |
| kimi | To determine if the given differential equation is the total derivative of the potential function, we need to check if the differential equation can be derived from the potential function by taking its total derivative. The potential function is given by: \[ \phi(t, y) = t^2 + ty^2 \] The total derivative of \(\phi\) with respect to \(t\) is: \[ d\phi = \frac{\partial \phi}{\partial t} dt + \frac{\partial \phi}{\partial y} dy \] Let's calculate the partial derivatives: \[ \frac{\partial \phi}{\partial t} = 2t + y^2 \] \[ \frac{\partial \phi}{\partial y} = 2ty \] Now, let's form the total derivative: \[ d\phi = (2t + y^2) dt + 2ty dy \] For the differential equation to be the total derivative of \(\phi\), it must be of the form: \[ \frac{dy}{dt} = \frac{\frac{\partial \phi}{\partial y}}{\frac{\partial \phi}{\partial t}} \] Substituting the partial derivatives into the equation, we get: \[ \frac{dy}{dt} = \frac{2ty}{2t + y^2} \] Now, let's compare this with the given differential equation: \[ 2tyy' + 2t + ty^2 = 0 \] Rearrange the given differential equation to isolate \(y'\): \[ 2tyy' = -2t - ty^2 \] \[ yy' = -\frac{2t}{t} - \frac{ty^2}{2t} \] \[ yy' = -2 - \frac{y^2}{2} \] \[ y' = -\frac{2}{t} - \frac{y^2}{2t} \] Simplify the expression: \[ y' = -\frac{2 + \frac{y^2}{2}}{t} \] Now, let's compare this with the total derivative form: \[ \frac{2ty}{2t + y^2} \] It is clear that the given differential equation does not match the form of the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\). Therefore, the given differential equation is not the total derivative of the potential function \(\phi(t, y)\). |
False |
| wenxinyiyan | 我们首先计算给定势函数$\phi(t, y) = t^2 + ty^2$的全导数。 全导数的计算公式为： $d\phi = \frac{\partial \phi}{\partial t}dt + \frac{\partial \phi}{\partial y}dy$对$\phi(t, y) = t^2 + ty^2$求偏导数，我们有： $\frac{\partial \phi}{\partial t} = 2t + y^2, \quad \frac{\partial \phi}{\partial y} = 2ty$因此，势函数$\phi(t, y)$的全导数为： $d\phi = (2t + y^2)dt + 2tydy$将全导数$d\phi$与给定的微分方程$2tyy' + 2t + ty^2 = 0$进行比较，我们可以看到它们并不相同。 因此，给定的微分方程$2tyy' + 2t + ty^2 = 0$不是势函数$\phi(t, y) = t^2 + ty^2$的全导数。 |
False |
| zhipu | To determine whether the differential equation \(2tyy' + 2t + ty^2 = 0\) is the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\), we need to check if the differential equation can be expressed in the form: \[ \frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial y} y' = 0. \] First, let's compute the partial derivatives of \(\phi(t, y)\): 1. The partial derivative of \(\phi\) with respect to \(t\): \[ \frac{\partial \phi}{\partial t} = \frac{\partial}{\partial t} (t^2 + ty^2) = 2t + y^2. \] 2. The partial derivative of \(\phi\) with respect to \(y\): \[ \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} (t^2 + ty^2) = 2ty. \] Now, we form the total derivative of \(\phi\) with respect to \(t\): \[ \frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial y} y' = (2t + y^2) + (2ty) y'. \] For the given differential equation \(2tyy' + 2t + ty^2 = 0\), we can rewrite it as: \[ 2tyy' + 2t + ty^2 = (2t + y^2) + (2ty) y'. \] This matches the form of the total derivative \(\frac{d\phi}{dt}\) that we computed above. Therefore, the given differential equation \(2tyy' + 2t + ty^2 = 0\) is indeed the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\). ### Conclusion: Yes, the differential equation \(2tyy' + 2t + ty^2 = 0\) is the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\). |
False |
| gpto1 | To determine whether the given differential equation is the total derivative of the potential function \(\phi(t, y) = t^2 + t y^2\), we'll compute the total derivative \(\frac{d\phi}{dt}\) along the solutions \(y(t)\) and compare it to the left-hand side of the differential equation. If they match, then the differential equation represents the total derivative of \(\phi\). Let's start by computing the total derivative \(\frac{d\phi}{dt}\): 1. **Compute the partial derivatives of \(\phi\):** \[ \frac{\partial \phi}{\partial t} = 2t + y^2, \quad \frac{\partial \phi}{\partial y} = 2 t y \] 2. **Compute the total derivative:** \[ \frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial y} \cdot y' = \left(2t + y^2\right) + \left(2 t y\right) y' = 2 t y y' + 2 t + y^2 \] Now, let's write the given differential equation: \[ 2 t y y' + 2 t + t y^2 = 0 \] Notice that in the expression for \(\frac{d\phi}{dt}\), the last term is \(y^2\), whereas in the differential equation, the last term is \(t y^2\). This discrepancy shows that the left-hand side of the differential equation is not exactly equal to \(\frac{d\phi}{dt}\). **Comparison:** \[ \frac{d\phi}{dt} = 2 t y y' + 2 t + y^2 \quad \text{(from total derivative)} \] \[ 2 t y y' + 2 t + t y^2 \quad \text{(from differential equation)} \] Subtracting the two expressions, we find: \[ \left(2 t y y' + 2 t + y^2\right) - \left(2 t y y' + 2 t + t y^2\right) = y^2 - t y^2 = y^2 (1 - t) \] Since this difference is not zero (unless \(t = 1\), which is not generally the case), the given differential equation does not represent the total derivative of \(\phi\). **Conclusion:** No, because when we compute the total derivative of \(\phi(t, y)\), it does not match the equation; thus, the differential equation is not the total derivative of \(\phi\). |
False |
| deepseekv3 | To determine if the given differential equation is the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\), we follow these steps: ### Step 1: Compute the partial derivatives of \(\phi(t, y)\) The potential function is: \[ \phi(t, y) = t^2 + ty^2 \] Compute its partial derivatives with respect to \(t\) and \(y\): \[ \frac{\partial \phi}{\partial t} = 2t + y^2 \] \[ \frac{\partial \phi}{\partial y} = 2ty \] ### Step 2: Compute the total derivative of \(\phi(t, y)\) The total derivative of \(\phi(t, y)\) with respect to \(t\) is: \[ \frac{d\phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial y} \cdot y' \] Substitute the partial derivatives: \[ \frac{d\phi}{dt} = 2t + y^2 + 2ty \cdot y' \] ### Step 3: Compare with the given differential equation The given differential equation is: \[ 2tyy' + 2t + ty^2 = 0 \] Rearrange the total derivative of \(\phi(t, y)\): \[ \frac{d\phi}{dt} = 2t + y^2 + 2tyy' \] This can be rewritten as: \[ \frac{d\phi}{dt} = 2t + 2tyy' + y^2 \] Notice that this matches the given differential equation: \[ 2tyy' + 2t + ty^2 = 0 \] Thus, the differential equation is indeed the total derivative of \(\phi(t, y)\). ### Conclusion Yes, the differential equation \(2tyy' + 2t + ty^2 = 0\) is the total derivative of the potential function \(\phi(t, y) = t^2 + ty^2\). |
False |