If dy/dx=(xy^2) and if y=1 when x=0, then when y=3, x is equal to
a) 1/8
b) - π/8
c) 1
d) (2√3)/3
I have answer D
To find the value of x when y = 3, we need to solve the given differential equation.
Given: dy/dx = xy^2
Separating the variables, we have:
dy/y^2 = x dx
Integrating both sides:
∫dy/y^2 = ∫x dx
Using the power rule of integration, the left-hand side becomes:
-1/y = (1/2) x^2 + C1
To find the constant of integration (C1), we can use the initial condition given:
When x = 0, y = 1
Substituting these values into the equation above:
-1/1 = (1/2) (0)^2 + C1
-1 = C1
Therefore, the equation becomes:
-1/y = (1/2) x^2 - 1
To solve for x when y = 3, substitute y = 3 into the equation:
-1/3 = (1/2) x^2 - 1
Multiplying both sides by 6 to eliminate fractions:
-2 = 3x^2 - 6
Rearranging the equation:
3x^2 = 4
Dividing both sides by 3:
x^2 = 4/3
Taking the square root of both sides:
x = ± √(4/3)
Since y = 3 and x is positive, we take the positive square root:
x = √(4/3)
Simplifying the square root:
x = √(4)/√(3)
x = 2/√(3)
x = (2√3)/3
Therefore, when y = 3, x is equal to (2√3)/3.
The correct answer is d) (2√3)/3.
To find the value of x when y=3, we need to solve the given differential equation:
dy/dx = xy^2
We can start by separating the variables and integrating both sides:
dy/y^2 = x dx
Integrating both sides gives:
-1/y = (x^2)/2 + C
Where C is the constant of integration. Now, we can use the initial condition y=1 when x=0 to find the value of C:
-1/1 = (0^2)/2 + C
-1 = C
Plugging this value of C back into our equation, we have:
-1/y = (x^2)/2 - 1
Now, we can substitute y=3 and solve for x:
-1/3 = (x^2)/2 - 1
Adding 1/3 to both sides gives:
0 = (x^2)/2 - 1 + 1/3
Simplifying further:
0 = (x^2)/2 + 2/3
Multiplying both sides by 2, we have:
0 = x^2 + 4/3
Now, we can solve for x by taking square roots:
x^2 = -4/3
Since the square root of a negative number is not real, we conclude that there is no real solution for x when y=3. Therefore, none of the given options (a, b, c, d) are correct.
dy/dx=(xy^2) and if y=1 when x=0, then when y=3, x
dy/dx = xy^2
dy/y^2 = x dx
1/3 y^3 = 1/2 x^2 + C/3
y^3 = 3/2 x^2 + C
1 = 3/2 + C
C = -1/2
y^3 = 1/2 (3x^2-1)
3^3 = 1/2 (3x-1)
54 = 3x-1
x = 55/3
I think your answers are for a different problem.