actually an ODE question(but there's no diff. eq. subforum)
(y^2)(dy/dx)+y^3/x =2/x^2
using substitution u(x)=y^3
so I've been doing a few of these and getting them all wrong and this problem... I don't know where to start.
I tried (1/u)(du/dx)+u/x(du/dx)=2/x^2
but that gives me (after moving u/x to the other side of the equation and resubstituting y in for u)
ln|y|+c=(2/x^2)-y/x
and I don't know how to deal with that.
I'm pretty sure I can't just plug u^(1/3) in for y, but I don't know why nor do I know what to do.
To solve this ordinary differential equation, you correctly made the substitution u(x) = y^3. Now let's see how to proceed from there.
First, we need to find the derivatives involved. Differentiating u(x) = y^3 with respect to x, we get:
du/dx = 3y^2 * (dy/dx) ...(1)
Next, we substitute this expression for du/dx and y = u^(1/3) back into the original equation:
(1/u)(du/dx) + (u/x)(du/dx) = 2/x^2
Substituting the expression for du/dx from equation (1):
(1/u)(3y^2 * (dy/dx)) + (u/x)(3y^2 * (dy/dx)) = 2/x^2
Now, we substitute y = u^(1/3):
(1/u)(3(u^(2/3))) + (u/x)(3(u^(2/3))) = 2/x^2
Simplifying, we have:
3(u^(1/3))/u + 3(u^(1/3))/x = 2/x^2
Multiplying through by u^(-1/3)x^2, we get:
3x^2/u^(1/3) + 3x/u^(1/3) = 2
Next, we multiply both sides by u^(1/3) to eliminate the fractional exponent:
3x^2 + 3xu^(1/3) = 2u^(1/3)
Now, notice that we have an equation in terms of u and x, without any derivatives involved. We can solve this equation for u(x). Let's rearrange it:
2u^(1/3) - 3xu^(1/3) = 3x^2
Factor out u^(1/3):
u^(1/3)(2 - 3x) = 3x^2
Now, isolate u^(1/3) by dividing both sides by (2 - 3x):
u^(1/3) = 3x^2 / (2 - 3x)
Finally, raise both sides to the power of 3 to solve for u(x):
u(x) = (3x^2 / (2 - 3x))^3
At this point, you have found the expression for u(x) in terms of x. To find y(x), we take the cube root of u(x):
y(x) = u(x)^(1/3)
Substitute the expression for u(x) into this equation to obtain y(x) in terms of x.