solve the initial value problem by separation of variables dy/dx=6x^2y and y(0)=4
dy/y = 6 x^2 dx
ln y = 2 x^3 + c
when x = 0, y = 4
ln 4 = c
so
ln y = 2 x^3 + ln 4
ln (y/4) = 2 x^3
or
y/4 = e^(2x^3)
y = 4 e^(2x^3)
damon is correct, I lost y.
Suppose day/dx=6x+4pick possible expression for Y
Why did the chicken cross the Möbius strip? To get to the same side! Now let's tackle your initial value problem.
To solve the initial value problem dy/dx = 6x^2y, we can separate the variables by moving all the y terms to the left-hand side and all the x terms to the right-hand side:
dy/y = 6x^2dx
Now let's integrate both sides:
∫(dy/y) = ∫(6x^2dx)
The integral of dy/y is just ln|y|, and the integral of 6x^2dx is 2x^3 + C, where C is the constant of integration. Applying this, we get:
ln|y| = 2x^3 + C
To get rid of the absolute value, we can take the exponent of both sides:
|y| = e^(2x^3 + C)
But remember, we have an initial condition y(0) = 4. So let's substitute x = 0 and y = 4 into the equation:
|4| = e^(2(0)^3 + C)
This simplifies to:
4 = e^C
Now, we can rewrite our equation as:
|y| = 4e^(2x^3)
But since we want to solve for y, we need to consider two cases: y = 4e^(2x^3) and y = -4e^(2x^3).
Thus, the solution to the initial value problem is:
y = ±4e^(2x^3)
And there you have it! The solution that popped out of my clown hat.
To solve the given initial value problem dy/dx = 6x^2y and y(0) = 4 by separation of variables, we will follow the steps outlined below:
Step 1: Separate the variables
Rewrite the equation in a way that separates the variables x and y. In this case, we can divide both sides of the equation by y to move dy to one side and dx to the other:
dy/y = 6x^2 dx
Step 2: Integrate both sides
Integrate both sides of the equation with respect to their respective variables. For the left side, integrate with respect to y, and for the right side, integrate with respect to x:
∫ (dy/y) = ∫ (6x^2 dx)
The integral of dy/y can be directly evaluated as ln|y|. The integral of 6x^2 dx can be found using the power rule for integration:
ln|y| = 2x^3 + C
Here, C represents the constant of integration.
Step 3: Solve for y
Using algebraic manipulations, solve for y:
ln|y| = 2x^3 + C
Taking the exponential of both sides, we get:
|y| = e^(2x^3 + C)
Applying the absolute value to both sides allows us to consider both positive and negative values of y:
y = ± e^(2x^3 + C)
Step 4: Apply the initial condition
Use the initial condition y(0) = 4 to find the value of the constant C.
We substitute x = 0 and y = 4 into the equation:
4 = ± e^(2(0)^3 + C)
4 = ± e^(C)
Since e^(C) is always positive, we can ignore the negative sign and rewrite the equation as:
4 = e^(C)
Taking the natural logarithm of both sides, we find:
ln(4) = C
Therefore, the constant of integration C is ln(4).
Step 5: Final solution
Replace the constant C with its value in the solution obtained in Step 3:
y = ± e^(2x^3 + ln(4))
Simplifying further:
y = ± 4e^(2x^3)
Hence, the final solution to the initial value problem dy/dx = 6x^2y, y(0) = 4 is:
y = ± 4e^(2x^3)
ydy=6x^2 dx
y^2/2=2x^3 + C
y(0)=4
16/2=C