Diff EQ- exponential raised to the exponential?

[SmashtheVan]

Homework Statement

Solve the equation:

$$(x+xy^{2})dx + e^{x^{2}}y dy =0$$

Homework Equations

n/a

The Attempt at a Solution

i) $$xdx(1+y^{2}) = -e^{x^{2}}y dy$$
(multiply both sides by 2 to prepare for integration:

ii) $$\frac{-2xdx}{e^{x^{2}}}= \frac{2y dy}{1+y^{2}}$$

integrate and get:
iii) $$\frac{1}{e^{x^{2}}}= ln(1+y^{2}) + C$$

first integral was acheived from Maple, I am not sure if its what should be used here, but its what i have for now

iv) exponentiate both sides to remove natural log:

$$e^{e^{-x^{2}}}=1 + y^{2} + C$$

v)combine constants, separate y

$$y^{2} = e^{e^{-x^{2}}} - C$$

vi)square root of the system:

$$y=\sqrt{ e^{e^{-x^{2}}}} -C$$Does this seem like a reasonable answer? I'm wary because of the occurance of the exponential function, raised to the exponential function.

 

[Mark44]

One thing that's nice about differential equations is that when you get a solution, you can check it by seeing if it satisfies the DE.

As a side note, you show the same C in your last three equations. Actually these are all different constants, plus the third equation does not follow from the second. sqrt(a + b) != sqrt(a) + sqrt(b).

 

[SmashtheVan]

As a side note, you show the same C in your last three equations. Actually these are all different constants, plus the third equation does not follow from the second. sqrt(a + b) != sqrt(a) + sqrt(b).

yea, I am still getting used to keeping track of my C's...thanks for that note

i don't know why i didnt think to substitute the solution back in. thanks!