: Can someone check my working on these Diff Eqs.

[samir9]

URGENT: Can someone check my working on these Diff Eqs.

Hi I have a maths exam tomorrow, and want to make sure I am doing the right thing, please bare in mind i study physics, and am not used to some of the techniques that are used in pure maths, so if the person can help using the methods i am using, would be much appreciated.


Question 1)
Solve this first order differential equation

dy/dx = x^3/y

subject to the boundary condition that y(x=0)=1


ok so what I do is rearrange to get:

1/y dy = x^3 dx

and then integrate both sides with respect to their variable to get

ln y = 1/4 x^4 + c

so then i remove the ln:

y = e^(1/4 x^4 + c)

SO
now I am a little bit confused with the boundary conditions, am I right in saying that when x = 0 , y = 1??

if that is true then am I right in saying that the constant of integration must be 0 (as e^0 = 1), so the answer is:

y = e ^ (1/4 x^4)

??

Question 2)
Solve the 2nd order differential equation:

d^2y/dx^2 - 4dy/dx = 0

(no boundary conditions this time)

ok so here i used the substitution U = dy/dx and get

dU/dx - 4U = 0

and rearrange to get:

1/U dU = 4 dx

and similarly to above integrate both sides

ln U = 4x + c1

rearrange:

dy/dx = U = e ^ (4x +c1)

ok, so now I am slightly confused, do I ignore the constant becasue there are no bc's in this question? If that is right, then I just integrate again to get:

y = 4 e^(4x) + C

is that correct??

 

[benorin]

Question #1 is correct: well-done.
Question #2 is correct, except that you need the constant of integration to remain, so your answer is y=\frac{1}{4}e^{4x+C_1}+C_2 and it should be 1/4 not 4. This may be simplified since e^{4x+C_1}=e^{4x}e^{C_1}=Ce^{4x},\, C>0

 

[samir9]

wow thanks for the quick reply, unfortunately I wasnt able to access the site to check for a reply yesterday..

question 1 is wrong though, look closely, the LHS isn't 1/y... Just realized myself, the exam went well anyways.


THANKSSSS!