1. (x+a) y' -3y = (x+a)^4
2. Interchange dependent and independent variables to solve:
(2x-5y^3) y' + y = 0, y(1)=1
3. Using the subst. z=y^-2 solve:
y' = xy - y^3 e^-x^2
4. Using the subst. 1+x = e^t solve:
(1+x)^2 y" + (1+x) y' + y =4cos(ln(1+x))
Any help will be much appreciated.
What you need to do here is rewrite the equation so the DE is in terms of z and x.
Since , we need to find . Applying the chain rule, we know that
Now make all these substitutions into the equation:
.
Multiplying both sides by , we get
This is a linear equation. Apply the integrating factor, where .
Can you take it from here?
--Chris
Thanks, I got the answers to be y=x(x+a)^3 + C(x+a)^3, y=1/(SQRT(2(x+C)e^-x^2)) and y=A(1+x)^i + B/(1+x)^i +4sin(ln(1+x)), for 1, 3, 4 respectively. Could someone be kind enough to check these answers for me please. But I'm still having problems with question 2, if someone could please help that would be much appreciated.
The solution to the DE does not contain non-real exponents.Originally Posted by Mr Fantastic
I suggest you review the general solution to a second order DE with constant coefficients, particularly the bit about the solutions to the auxillary equation being non-real.
The homogeneous solution to is .