Animated Solution for Mathematics - Differential Equations: Let y=y(x) be the solution of the differential equation ex1−y2dx+(xy)dy=0,y(1)=−1. Then the value of (y(3))2 is equal to:
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Visualized Solution
Analyzing the Differential Equation
Given differential equation: ex1−y2dx+(xy)dy=0
Initial condition: y(1)=−1
Goal: Find the value of (y(3))2
Separating the Variables
Rearrange the equation to isolate terms:
xydy=−ex1−y2dx
Divide by 1−y2 and multiply by x to separate variables:
1−y2ydy=−xexdx
Setting up the Integrals
Integrate both sides of the equation:
∫1−y2ydy=−∫xexdx
Integrating the y-side
To solve ∫1−y2ydy, use substitution:
Let 1−y2=u⇒−2ydy=du⇒ydy=−21du
The integral becomes: ∫−2u1du=−u
Substituting back u: −1−y2
Integrating the x-side
To solve −∫xexdx, use integration by parts:
Let u=x and v=ex
−[xex−∫1⋅exdx]
=−[xex−ex]=ex(1−x)
Combining the Results
Combine the integrated sides and add the constant of integration C:
−1−y2=ex(1−x)+C
Applying Initial Conditions
Use the initial condition y(1)=−1 to find C:
Substitute x=1 and y=−1 into the equation:
−1−(−1)2=e1(1−1)+C
Finding the Constant C
Simplify the expression:
−1−1=e(0)+C
0=0+C⇒C=0
The Particular Solution
Substitute C=0 back into the general equation:
−1−y2=ex(1−x)
Multiply by −1 to simplify:
1−y2=ex(x−1)
Evaluating at x=3
Substitute x=3 into the particular solution:
1−(y(3))2=e3(3−1)
1−(y(3))2=2e3
Final Calculation for (y(3))2
Square both sides to solve for (y(3))2:
1−(y(3))2=(2e3)2
1−(y(3))2=4e6
Rearrange to solve:
(y(3))2=1−4e6
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The Sigma Insight: Variable Separable Method
Analyzing the Setup
Welcome, future engineers and mathematicians! Today, we are going to embark on a journey through a beautiful differential equation.
Often, when students see an equation like ex1−y2dx+(xy)dy=0, they feel a sense of dread. They see mixed variables, square roots, and exponential functions, but this is not a monster; it is a puzzle waiting for you to reveal its hidden order.
The Art of Separation
The first step is the separation of variables. We start with the given equation:
ex1−y2dx+(xy)dy=0
Our goal is to isolate dy with y and dx with x. By rearranging, we get:
xydy=−ex1−y2dx
Now, we divide both sides by 1−y2 and multiply by x to obtain the elegant form:
1−y2ydy=−xexdx
The variables are now perfectly separated, with the left side being a function of y and the right side a function of x.
The Integration Challenge
Now that we have separated the variables, we must integrate both sides:
∫1−y2ydy=−∫xexdx
For the left side, we use substitution. Let u=1−y2, then du=−2ydy, which implies ydy=−21du. Substituting this into the integral, we get:
−21∫u−1/2du=−u=−1−y2
For the right side, we use integration by parts with u=x and dv=exdx. Applying the formula ∫udv=uv−∫vdu, we get:
−∫xexdx=−(xex−∫exdx)=−(xex−ex)=ex(1−x)
The Constant of Integration
Combining our results, we obtain the general solution:
−1−y2=ex(1−x)+C
We are given the initial condition that the curve passes through the point (1,−1). Substituting x=1 and y=−1 into our equation:
−1−(−1)2=e1(1−1)+C⇒0=0+C⇒C=0
The constant vanishes, leaving us with the particular solution:
1−y2=ex(x−1)
Final Calculation
We are at the finish line. We need to find (y(3))2. Substituting x=3 into our particular solution:
1−(y(3))2=e3(3−1)=2e3
To isolate (y(3))2, we square both sides:
1−(y(3))2=(2e3)2=4e6
Rearranging the terms, we find the final result:
(y(3))2=1−4e6
You have mastered the separation, the integration, and the application of initial conditions. Keep this confidence with you as you tackle the next problem!