Question

Let \(u\) and \(v\) be continuously differentiable with respect to \(x\) and satisfy $$ \begin{aligned} u+2 u^{2}+v^{2}+x^{2}+2 v-x &=0 \\ x u v+e^{u} \sin (v+x) &=0 \end{aligned} $$ and \(u(0)=v(0)=0 .\) Find \(u^{\prime}(0)\) and \(v^{\prime}(0)\).

Short answer

Answer: u'(0) = 0 and v'(0) = 1/2.

Step-by-step solution

Partial derivatives of the given equations

To find the derivatives of u(x) and v(x) with respect to x, first, differentiate the given equations with respect to x using the chain rule: \(\frac{d}{d x}\left(u+2 u^{2}+v^{2}+x^{2}+2 v-x\right) = \frac{d(0)}{d x}\) and \(\frac{d}{d x}\left(x u v+e^{u} \sin (v+x)\right) = \frac{d(0)}{d x}\) After differentiating both sides of the equations with respect to x, we have the following expressions: $$ \begin{aligned} u^{\prime}+4 uu^{\prime}+2 v v^{\prime}+2 x+2 v^{\prime}-1 &=0\\ u v+x (uv^{\prime}+u^{\prime} v)+ e^{u}u^{\prime}\sin(v+x)+e^u \cos(v+x)(v^{\prime}+1) &= 0 \end{aligned} $$

Solving the equations

We can now solve for u'(x) and v'(x) by the implicit differentiation method. Rewriting the first equation for u'(x): $$u^{\prime}(1+4u) = -2v v^{\prime}(x) -2x -2 v^{\prime}(x) +1$$ And from the second equation, we can solve for u'(x): $$u^{\prime}(x)(u v e^u\sin{(v+x)} + v e^u \cos{(v+x)} - x v) = xv^{\prime} e^u \cos{(v+x)} - e^u \cos{(v+x)}(v^{\prime}+1)$$ We have now re-arranged both equations to solve for u'(x) with respect to v'(x).

Finding u'(0) and v'(0)

Since we are given the initial conditions u(0)=v(0)=0, we can plug x=0 into the equations we derived in step 2, which will simplify them significantly: For u'(0), $$u^{\prime}(0)(1 + 4u(0)) = -2v(0) v^{\prime}(0) -2(0) - 2v^{\prime}(0) + 1$$ $$0 = -2v^{\prime}(0)+1$$ Solving for v'(0): $$v^{\prime}(0) = \frac{1}{2}$$ Next, for u'(0), $$u^{\prime}(0)(u(0) v(0) e^{u(0)}\sin{(v(0) + 0)} + v(0) e^{u(0)} \cos{(v(0) + 0)} - 0 v(0)) = 0 v^{\prime}(0) e^{u(0)} \cos{(v(0) + 0)} - e^{u(0)} \cos{(v(0) + 0)}(v^{\prime}(0)+1)$$ This simplifies to: $$0 = - e^0 \cos{(0 + 0)}(\frac{1}{2}+1)$$ Which gives us the value of u'(0): $$u^{\prime}(0) = 0$$ In conclusion, we have found the values of \(u^{\prime}(0) = 0\) and \(v^{\prime}(0) = \frac{1}{2}\).

Key concepts

Continuously Differentiable

The term "continuously differentiable" refers to functions that possess derivatives which are continuous. This means, for a function like \(u(x)\) or \(v(x)\), not only must the derivative exist at every point, but it should also be a smooth and unbroken curve. A function being continuously differentiable guarantees that its rate of change is predictable and consistent.
When a problem states that \(u\) and \(v\) are continuously differentiable with respect to \(x\), it implies all derivatives up to the first order exist without abrupt changes. Such conditions are critical when performing tasks like implicit differentiation, as they ensure the derivatives taken at any point are valid.
This smoothness comes in handy while dealing with differentiability over intervals. No jumps or sharp turns in the graph of the function lead to reliable calculations. It's a fundamental assumption in many calculus problems involving derivatives and can simplify our approach to solving them.

Chain Rule

The chain rule is an essential formula in calculus used to differentiate composite functions. A composite function is when one function is placed inside another, like \(h(x) = f(g(x))\). If we need to find the derivative of such functions, the chain rule is our tool.
Simply put, the chain rule states: If \(y = f(u)\) and \(u = g(x)\), then the derivative \(\frac{dy}{dx}\) is given by \(\frac{dy}{du} \times \frac{du}{dx}\). This rule allows us to "chain" together the derivatives of the inner and outer functions.
In the exercise, the equations involving \(u\) and \(v\) utilize the chain rule when differentiating terms like \(e^u \sin(v+x)\) with respect to \(x\).

• First, differentiate the outer function, keep the inner function as it is.
• Then differentiate the inner function.

This step-by-step differentiation is crucial for implicitly differentiating and solving equations to find \(u'(0)\) and \(v'(0)\).

Initial Conditions

Initial conditions are values given to a function at a specific point, often used to find particular solutions to differential equations. In this problem, we have the initial conditions \(u(0) = 0\) and \(v(0) = 0\). Initial conditions allow us to determine specific coefficients and constants in solutions, which would otherwise remain general.
When performing implicit differentiation, initial conditions play a pivotal role in finding particular derivatives at a given point. By substituting these conditions into the derived expressions, we can simplify the equations significantly. For instance, setting \(x = 0\), \(u(0) = 0\), and \(v(0) = 0\) allows us to solve for \(u'(0)\) and \(v'(0)\) without the added complication of unknown variables.
Initial conditions provide a starting point or boundary that can translate general solutions into specific numerical answers, making them invaluable tools in calculus, particularly in solving real-world problems modeled by differential equations.