# Nathan Curtis' Question at Yahoo! Answers regarding Differential Equations

#### Chris L T521

##### Well-known member
Staff member
Here is the question:

Nathan Curtis said:
Let I be an open interval containing 1 and f : I → R be a function such that f(1) = 2. Assume f is a solution of the diﬀerential equation y′ − y^2=x^2 . Find f'(1) f''(1) and f'''(1).
Here is a link to the question:

I have posted a link there to this topic so the OP can find my response.

#### Chris L T521

##### Well-known member
Staff member
Hi Nathan Curtis,

In this problem, we're assuming that $y=f(x)$ is a solution to the differential equation $y^{\prime}-y^2 = x^2$. Thus, in terms of $f(x)$, this means that
$f^{\prime}(x) - (f(x))^2 = x^2 \implies f^{\prime}(x) = (f(x))^2 + x^2.$
Since we know that $f(1)=2$, we now see that
$f^{\prime}(1) = (f(1))^2 + (1)^2 = 2^2+1^2 = 5.$
Differentiating the expression for $f^{\prime}(x)$ yields the equation
$f^{\prime\prime}(x) = 2 f(x)\cdot f^{\prime}(x) + 2x$
and thus
$f^{\prime\prime}(1) = 2 f(1)\cdot f^{\prime}(1) + 2(1) = 2(2)(5) + 2 = 22.$
Differentiating the expression for $f^{\prime\prime}(x)$ yields the equation
$f^{\prime\prime\prime}(x) = 2 (f^{\prime}(x))^2 + 2 f(x)\cdot f^{\prime\prime}(x) + 2$
and thus
$f^{\prime\prime\prime}(1) = 2 (f^{\prime}(1))^2 + 2 f(1)\cdot f^{\prime\prime}(1) + 2 = 2(5)^2+2(2)(22) + 2 = 140$

Therefore, if $f(1)=2$, then $f^{\prime}(1)=5$, $f^{\prime\prime}(1) = 22$ and $f^{\prime\prime\prime}(1) = 140$.

I hope this makes sense!

#### nathancurtis11

##### New member
Thank you so much for the help and linking me to this website! I'm sure I will be using it a lot this semester because I can hardly understand my professor in diff eq. This helped a lot though and made me realize where I made my original mistake. Again thanks so much!