I got y = e^(x^3/3) + 2.
Damon
answered
dy/dx = x^2 e^(x^3/3)
but
e^(x^3/3) = y-2
so
dy/dx = x^2 (y-2) hummm
I'll try
dy/(y-1) = x^2 dx
ln(y-1)= x^3/3 + c
(y-1) = e^[x^3/3 + c]
y-1 = e^(x^3/3) e^c
y-1 = Ce^(x^3/3)
y = 1 + Ce^(x^3/3)
3 = 1 + C
C = 2
so
y = 1 + 2e^(x^3/3)
Explain Bot
answered
Starting with the given differential equation: dy/dx = x^2(y - 1)
Separating the variables by bringing dy to one side and dx to the other side: (1/(y - 1))dy = x^2 dx
Now, we can integrate both sides. Integrating the left side involves using the natural logarithm function (ln), while integrating the right side involves using the power rule for integration.
∫(1/(y - 1))dy = ∫x^2 dx
To integrate 1/(y - 1) with respect to y, we can use the substitution u = y - 1. This gives us du = dy, and the integral becomes:
∫(1/u)du = ln|u|
Substituting back u = y - 1, we have:
ln|y - 1| = ∫x^2 dx
Integrating x^2 with respect to x gives us:
ln|y - 1| = (1/3)x^3 + C
Where C is the constant of integration.
Next, we can solve for y by exponentiating both sides:
|y - 1| = e^((1/3)x^3 + C)
To simplify the equation, we can remove the absolute value by considering two cases:
Case 1: y - 1 > 0, implies y > 1
In this case, we have:
y - 1 = e^((1/3)x^3 + C)
y = e^((1/3)x^3 + C) + 1
Case 2: y - 1 < 0, implies y < 1
In this case, we have:
-(y - 1) = e^((1/3)x^3 + C)
-y + 1 = e^((1/3)x^3 + C)
y - 1 = -e^((1/3)x^3 + C)
y = -e^((1/3)x^3 + C) + 1
So, the particular solution to the given differential equation with initial condition f(0) = 3 is:
y = e^(x^3/3) + 1 (since f(0) = 3 implies y > 1)
Step-by-Step Bot
answered
Step 1: Separate the variables.
Rewrite the given differential equation as:
dy / (y - 1) = x^2 dx.
Step 2: Integrate both sides.
Integrating both sides, we get:
∫ dy / (y - 1) = ∫ x^2 dx.
Step 3: Integrate the left side.
The integral of 1 / (y - 1) with respect to y is ln|y - 1|.
Therefore, the left side becomes ln|y - 1| + C1, where C1 is the constant of integration.
Step 4: Integrate the right side.
The integral of x^2 with respect to x is (1/3) x^3.
Therefore, the right side becomes (1/3) x^3 + C2, where C2 is the constant of integration.
Step 5: Apply the initial condition.
Substitute the initial condition f(0) = 3 into the equation.
ln|3 - 1| + C1 = (1/3) (0)^3 + C2.
Simplifying this equation, we have:
ln(2) + C1 = C2.
Step 6: Combine the constants.
Combine the constants C1 and C2 into a single constant C:
ln(2) + C1 = C.
Step 7: Solve for y.
Substituting the constants back into the equation, we get:
ln|y - 1| = (1/3) x^3 + ln(2) + C.
Exponentiating both sides, we have:
|y - 1| = e^((1/3) x^3 + ln(2) + C).
Since the absolute value can be positive or negative, we have two cases to consider:
Case 1: y - 1 > 0:
In this case, we can remove the absolute value:
y - 1 = e^((1/3) x^3 + ln(2) + C).
Case 2: y - 1 < 0:
In this case, we need to flip the sign:
-(y - 1) = e^((1/3) x^3 + ln(2) + C).
Simplifying, we get:
1 - y = e^((1/3) x^3 + ln(2) + C).
Step 8: Find y.
For Case 1: y - 1 = e^((1/3) x^3 + ln(2) + C).
Adding 1 to both sides, we have:
y = 1 + e^((1/3) x^3 + ln(2) + C).
For Case 2: 1 - y = e^((1/3) x^3 + ln(2) + C).
Subtracting 1 from both sides, we get:
-y = -1 + e^((1/3) x^3 + ln(2) + C).
Multiplying both sides by -1, we obtain:
y = 1 - e^((1/3) x^3 + ln(2) + C).
Therefore, the particular solution to the given differential equation with the initial condition f(0) = 3 is:
y = 1 + e^((1/3) x^3 + ln(2) + C).
