If 𝑦(𝑥) is the solution of the differential equation
𝑥𝑑𝑦 − (𝑦2 − 4𝑦)𝑑𝑥 = 0 for 𝑥 > 0, 𝑦(1) = 2,
and the slope of the curve 𝑦 = 𝑦(𝑥) is never zero, then the value of 10 𝑦(√2 ) is
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Let's solve this differential equation. We're given that x dy minus y squared minus 4y dx equals 0, with the initial condition y of 1 equals 2, and x is greater than 0. First, we'll rewrite this in standard form by dividing both sides by x dx. This gives us dy/dx equals (y squared minus 4y)/x. To solve this, we'll separate the variables. We get dy over y squared minus 4y equals dx over x. Now we can integrate both sides. The vector field shows the direction of the solution curve at each point, with the initial condition at the point (1, 2).
Now we need to integrate both sides of our equation. Let's focus on the left side first. We have the integral of dy over y squared minus 4y. To integrate this, we'll use partial fraction decomposition. We can rewrite y squared minus 4y as y times y minus 4. So our fraction becomes 1 over y times y minus 4. Using partial fractions, we write this as A over y plus B over y minus 4. To find the constants A and B, we set up the equation 1 equals A times y minus 4 plus B times y. When y equals 0, we get 1 equals A times negative 4, so A equals negative 1/4. When y equals 4, we get 1 equals B times 4, so B equals 1/4. Therefore, our partial fraction decomposition is 1/4 times negative 1/y plus 1 over y minus 4. The graph shows the function with vertical asymptotes at y equals 0 and y equals 4.
Now we can integrate both sides of our equation. On the left side, we have the integral of 1/4 times negative 1/y plus 1 over y minus 4 with respect to y. This gives us 1/4 times negative natural log of absolute y plus natural log of absolute y minus 4. On the right side, we have the integral of dx over x, which is natural log of absolute x plus a constant C. Setting these equal, we get 1/4 times natural log of the absolute value of y minus 4 over y equals natural log of absolute x plus C. Using properties of logarithms, we can rewrite this as natural log of the absolute value of y minus 4 over y equals 4 times natural log of absolute x plus 4C. This simplifies to natural log of the absolute value of y minus 4 over y equals natural log of x to the 4th power plus natural log of K, where K is e to the 4C. Taking the exponential of both sides, we get the absolute value of y minus 4 over y equals K times x to the 4th power. Since x is positive, we can write this as y minus 4 over y equals K x to the 4th power, where K is a non-zero constant. The graph shows various solution curves for different values of K, with the curve for K equals negative 1 highlighted in red.
Now we'll find the particular solution by using the initial condition y of 1 equals 2. From our general solution, we have y minus 4 over y equals K x to the 4th power. We can rewrite this as 1 minus 4 over y equals K x to the 4th power. Substituting the initial condition, we get 1 minus 4 over 2 equals K times 1 to the 4th power. This simplifies to 1 minus 2 equals K, so K equals negative 1. Substituting this value back into our equation, we get 1 minus 4 over y equals negative x to the 4th power. Rearranging, we get 4 over y equals 1 plus x to the 4th power, which gives us our particular solution: y equals 4 over 1 plus x to the 4th power. The graph shows this solution curve in red, with the initial point at (1, 2) and another key point at (root 2, 4/5). The curve approaches the horizontal asymptote y equals 4 as x approaches 0, and approaches y equals 0 as x approaches infinity.
Now we can find the final answer. We need to calculate 10 times y of root 2. From our solution, y equals 4 over 1 plus x to the 4th power. Substituting x equals root 2, we get y of root 2 equals 4 over 1 plus root 2 to the 4th power. We know that root 2 to the 4th power equals 2 squared, which is 4. So y of root 2 equals 4 over 1 plus 4, which is 4 over 5. Therefore, 10 times y of root 2 equals 10 times 4 over 5, which equals 40 over 5, or 8. Let's also verify that the slope of our solution curve is never zero, as stated in the problem. The slope is given by dy/dx equals y squared minus 4y over x. For our solution, y is always between 0 and 4, so y squared minus 4y is always negative for x greater than 0. Since x is positive, the slope is never zero. Therefore, our final answer is 8.