Consider the differential equation dy/dx=-2x/y
Find the particular solution y =f(x) to the given differential equation with the initial condition f(1) = -1
Why did the mathematician bring a ladder to the differential equation party?
Because he wanted to find a solution that would "step up" his game!
Now, let's solve the equation. We'll start by separating the variables.
dy/y = -2x dx
Integrating both sides, we get:
ln|y| = -x^2 + C
To find the value of C, we can use the initial condition f(1) = -1.
So, plugging in x = 1 and y = -1, we have:
ln|-1| = -1^2 + C
ln(1) = -1 + C
0 = -1 + C
Therefore, C = 1.
Substituting this value back into the equation, we have:
ln|y| = -x^2 + 1
To find y, we can exponentiate both sides:
|y| = e^(-x^2 + 1)
Since we're looking for a particular solution, we can ignore the absolute value sign, giving us:
y = ±e^(-x^2 + 1)
Therefore, the particular solution to the given differential equation with the initial condition f(1) = -1 is:
y = ±e^(-x^2 + 1)
To find the particular solution y = f(x) to the differential equation dy/dx = -2x/y with the initial condition f(1) = -1, you can follow these steps:
Step 1: Separate the variables by multiplying both sides of the equation by y and dx:
y dy = -2x dx
Step 2: Integrate both sides with respect to their respective variables:
∫ y dy = ∫ -2x dx
Step 3: Integrate the left side:
(1/2) y^2 = -x^2 + C1
Step 4: Solve for y^2:
y^2 = -2x^2 + 2C1
Step 5: Apply the initial condition f(1) = -1 to find the particular value of the constant:
(-1)^2 = -2(1)^2 + 2C1
1 = -2 + 2C1
2 = 2C1
C1 = 1
Step 6: Substitute the value of C1 back into the equation:
y^2 = -2x^2 + 2(1)
y^2 = -2x^2 + 2
Step 7: Take the square root of both sides to get y:
y = ± √(-2x^2 + 2)
Thus, the particular solution to the given differential equation with the initial condition f(1) = -1 is y = ± √(-2x^2 + 2).
To find the particular solution y = f(x) to the given differential equation dy/dx = -2x/y with the initial condition f(1) = -1, we can use the method of separation of variables.
1. Start by rewriting the equation in the form:
y dy = -2x dx
2. Integrate both sides of the equation separately with respect to y and x:
∫ y dy = -∫ 2x dx
3. On the left side, integrate with respect to y:
(1/2) y^2 = -∫ 2x dx
4. On the right side, integrate with respect to x:
(1/2) y^2 = -x^2 + C
5. Set up the initial condition f(1) = -1 to find the value of the constant C:
(1/2) (-1)^2 = -1^2 + C
1/2 = -1 + C
C = 1/2 - (-1)
C = 1/2 + 1
C = 3/2
6. Substitute C = 3/2 back into the equation:
(1/2) y^2 = -x^2 + 3/2
7. Solve for y:
y^2 = -2x^2 + 3
y = ±√(-2x^2 + 3)
Therefore, the particular solution to the given differential equation with the initial condition f(1) = -1 is:
y = ±√(-2x^2 + 3)
dy/dx = -2x/y
y dy = -2x dx
1/2 y^2 = -x^2 + C
since y(1) = 1
-1+C = 1/2
C = 3/2
y^2 = -2x^2 + 3