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

We begin with $y^{\prime }+x^{3}y=3x^3,y(0)=8$. 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

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

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

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

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

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

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

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

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