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Chapter 4: Problem 14

\(x \sqrt{1-y^{2}} d x+y \sqrt{1-x^{2}} d y=0, y(0)=1\).

Short Answer

Expert verified
Answer: The implicit solution for the given exact differential equation is \(x^2\ \frac{\sqrt{1-y^2}}{2} + y^2\ \frac{\sqrt{1-x^2}}{2}- \frac{1}{2} = 0\).

Step by step solution

01

Check if the given differential equation is exact

To check if the given exact differential equation is exact, first calculate the partial derivatives: \(\frac{\partial}{\partial y}( x \sqrt{1-y^{2}} )= -x y \ \frac{1}{\sqrt{1-y^{2}}}\) \(\frac{\partial}{\partial x}( y \sqrt{1-x^{2}} )= -x y \ \frac{1}{\sqrt{1-x^{2}}}\) Since \(\frac{\partial}{\partial y}( x \sqrt{1-y^{2}} ) = \frac{\partial}{\partial x}( y \sqrt{1-x^{2}} )\), the differential equation is exact.
02

Identify the potential function 蠄(x, y)

To find the potential function 蠄(x, y) that satisfies the exact differential equation, we need to integrate the given expressions with respect to their respective variables: Integrate with respect to x: \(\int x \sqrt{1-y^{2}}dx = x^2\ \frac{\sqrt{1-y^2}}{2} + C_1(y)\) Integrate with respect to y: \(\int y \sqrt{1-x^{2}}dy = y^2\ \frac{\sqrt{1-x^2}}{2} + C_2(x)\)
03

Determine the constants of integration

Now, we have \(\psi(x, y) = x^2\ \frac{\sqrt{1-y^2}}{2} + y^2\ \frac{\sqrt{1-x^2}}{2} + C(x, y)\) The potential function, however, has no constants of integration, so we simply have: \(\psi(x, y) = x^2\ \frac{\sqrt{1-y^2}}{2} + y^2\ \frac{\sqrt{1-x^2}}{2}\)
04

Apply the initial condition

Now we apply the given initial condition \(y(0) = 1\): \(\psi(0, 1) = 0^2\ \frac{\sqrt{1-1^2}}{2} + 1^2\ \frac{\sqrt{1-0^2}}{2}\) \(\psi(0, 1) = 0 + \frac{1}{2}\) Which gives us: \(\psi(x, y) - \frac{1}{2} = 0\).
05

Write down the final implicit solution

Finally, we can rewrite the above expression for the implicit solution of the exact differential equation: \(x^2\ \frac{\sqrt{1-y^2}}{2} + y^2\ \frac{\sqrt{1-x^2}}{2}- \frac{1}{2} = 0\) This is the implicit solution to the given exact differential equation, considering the initial condition \(y(0) = 1\).

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Most popular questions from this chapter

Show that \(y=x-x^{-1}\) is a solution of the differential equation \(x y^{\prime}+y=2 x\)

Solve the following differential equations: (i) \(2 x y^{\prime}\left(x-y^{2}\right)+y^{3}=0\) (ii) \(4 y^{6}+x^{3}=6 x y^{5} y^{\prime}\) (iii) \(y\left(1+\sqrt{x^{2} y^{4}+1}\right) d x+2 x d y=0\) (iv) \(\left(\mathrm{x}+\mathrm{y}^{3}\right) \mathrm{dx}+3\left(\mathrm{y}^{3}-\mathrm{x}\right) \mathrm{y}^{2} \mathrm{~d} \mathrm{y}=0\)

A tank initially contains 50 litres of fresh water. Brine contains \(2 \mathrm{~kg}\) per litre of salt, flows into the tank at the rate of 2 litre per minutes and the mixture kept uniform by stirring runs out at the same rate. How long will it take for the quantity of salt in the tank to increase from 40 to \(80 \mathrm{~kg}\).

Solve the following differential equations: (i) \(\frac{d y}{d x}=\frac{x+y+1}{x+y-1}\) (ii) \(\left(\frac{x+y-1}{x+y-2}\right) \frac{d y}{d x}=\frac{x+y+1}{x+y+2}\)

A cup of tea is prepared in a preheated cup with hot water so that the temperature of both the cup and the brewing tea is initially \(190^{\circ} \mathrm{F}\). The cup is then left to cool in a room kept at a constant \(72^{\circ} \mathrm{F}\). Two minutes later, the temperature of the tea is \(150^{\circ} \mathrm{F}\). Determine (a) the temperature of the tea after 5 minutes. (b) the time required for the tea to reach \(100^{\circ} \mathrm{F}\).
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