# Help with solving linear differential equations using an integrating factor

I'm trying to solve this equation for $x=1$ using an integrating factor however for some reason my answer keeps turning out different from the book

In part a) I'm asked to find out $y$ using Euler's method when $x = 0.2$, my answer is correct here as $y = 4.46$.

The following equation is the one I'm trying to solve for $x=1$

$$\frac{dy}{dx} - 4xy = e^{2x^2}$$

In using taking $-4x$ and turning into the integrating factor I get

$$I(x) = e^{-2x^2}$$ Thus the equation becomes the following after multiplying through by the integrating factor $$\frac{d}{dx}(e^{-2x^2}y) = 1$$ So in following that line of logic I integrate both sides and try to find the value of $c$ using my knowledge from part a) but I keep coming up with the wrong answer

The integrated version for clarity

$$e^{-2x^2}y = x + c$$

Edit: I am given the initial condition $y=4$, $x=0$. I keep getting an answer in the region of $36$. My book is getting an answer of $x=1$, $y=4.55$.

• What is the Initial Condition here? – R. Singh May 8 '17 at 9:02
• initial condition $y =4$, $x=0$ – CertainlyNotAdrian May 8 '17 at 9:09
• What is the answer at $x=1$ according to your book? – R. Singh May 8 '17 at 9:20
• Re the suggestion you say you found in your book, note that if $y'(x)=e^{2x^2}+4xy(x)$ and $y(0)\geqslant0$ then $y'(x)\geqslant1$ for every $x\geqslant0$, thus, $y(x)\geqslant y(0)+x$ for every $x\geqslant0$ hence, if $y(0)=4$ then $y(1)\geqslant5$, and the condition that $y(1)=4.55$ is absurd. – Did May 8 '17 at 10:17
• Hence: typo (either in your book or in your transcription of your book). – Did May 8 '17 at 10:40

Your book's solution of $y(1)=4.55$ is certainly incorrect.
Based on the initial condition $y(0)=4$, one can evaluate the value of $c$. We have: $$e^{-2x}y=x+c$$ $$e^{0}\cdot 4=0+c$$ $$c=4$$ Therefore, we can deduce that the solution satisfying the initial condition is: $$y=xe^{2x}+4e^{2x}$$ Thus, the value of $y$ at $x=1$ should be: $$y(1)=1e^{2}+4e^{2}=5e^2\approx 36.94528049$$ Since you seem to be getting something "In the region of $36$", this is similar, if not exactly what you obtained.