Science:Math Exam Resources/Courses/MATH215/December 2013/Question 02 (a)/Solution 1

We begin with $y'+x^{3}y=3x^{3},\quad y(0)=8$. We can use an integrating factor here, taking $\mu (x)=e^{\int x^{3}dx}=e^{\frac {x^{4}}{4}}$ where we chose the arbitrary constant to be 0. Multiplying both sides of the equation by $\mu (x)$ gives

$e^{\frac {x^{4}}{4}}y'+e^{\frac {x^{4}}{4}}x^{3}y=3x^{3}e^{\frac {x^{4}}{4}}$ or

$(e^{\frac {x^{4}}{4}}y)'=3x^{3}e^{\frac {x^{4}}{4}}$.

We can integrate both sides of the equation (doing a simple substitution on the right-hand side) giving us

$\int {\frac {d}{dx}}(e^{\frac {x^{4}}{4}}y(x))dx=\int 3x^{3}e^{\frac {x^{4}}{4}}dx$

$e^{\frac {x^{4}}{4}}y=\underbrace {\int 3e^{u}du} _{u=x^{4}/4,\,du=x^{3}dx}=3e^{u}+C=3e^{\frac {x^{4}}{4}}+C$

$y=3+Ce^{\frac {-x^{4}}{4}}$.

If $y(0)=8$ then $8=3+C$ so $C=5$ and the final answer is

$y(x)=3+5e^{\frac {-x^{4}}{4}}$.