Find the solution to the differential equation dy/dx = (x^3)/(y^2),
where y(2) = 3
I have:
The integral of: dy(y^2) = the integral of: dx(x^3)
(y^3)/3 = (x^4)/4
The answer is supposed to come out to be y= the cube root of: ( 3/4x + 15 )
Thanks
The general solution is
(y^3)/3 = (x^4)/4 + C
where C is any constant. If y = 3 when x=2, then
27/3 = 16/4 + C
C = 9 - 4 = 5
y^3 = (3/4) x^4 + 3C = (3/4)x^4 + 15
y = cube root of [(3/4)x^4 + 15]
I believe you left out an exponent ^4 in the answer.
To find the solution to the given differential equation, we can separate the variables and integrate both sides. Here's how we can solve it step by step:
1. Start with the given differential equation: dy/dx = (x^3)/(y^2).
2. Rearrange the equation by multiplying both sides by y^2 and dx: y^2 dy = x^3 dx.
3. Now, let's integrate both sides of the equation. Integrate the left side with respect to y and the right side with respect to x.
∫y^2 dy = ∫x^3 dx
4. Evaluate the integrals:
(y^3)/3 = (x^4)/4 + C
Here, C is the constant of integration.
5. Now, to find the particular solution that satisfies the initial condition y(2) = 3, substitute the values into the equation.
(3^3)/3 = (2^4)/4 + C
Simplifying further:
27/3 = 16/4 + C
9 = 4 + C
C = 5
6. Substitute the value of C back into the general solution:
(y^3)/3 = (x^4)/4 + 5
7. Finally, solve the equation for y:
y^3 = (3/4)x^4 + 15
Taking the cube root of both sides, we get:
y = ∛[(3/4)x^4 + 15]
So, the solution to the given differential equation with the initial condition y(2) = 3 is y = ∛[(3/4)x^4 + 15].
Please note that in your answer, you missed the exponent ^4 in (3/4)x^4, so it should be y = ∛[(3/4)x^4 + 15], not y = ∛[(3/4)x + 15].