Question

Solve the following initial value problem (IVP);

(a) dy = 9.8 − 0.196y, y(0) = 48 by using the method of integrating factors.

Answer 1

We have the following DE:

$\cfrac{dy}{dx}=9.8-0.196y \\ \implies \frac{dy}{dx}+0.196y=9.8 \iff \frac{dy}{dx}+P_{(x)}y=Q_{(x)}$

Then we find the integrating factor as:

$\mu_{(x)}=e^ {\int { P_{(x)} {dx}}}=e^ {0.196 \int {{dx}}}=e^ {0.196x}$

We multiply the integrating factor for all the terms on the DE and we proceed to find y:

$\\ (e^ {0.196x} ) \frac{dy}{dx}+(0.196e^ {0.196x}) y=(9.8) e^ {0.196x} \\ \cfrac{d}{dx} \Big( y\cdot e^ {0.196x} \Big)=(9.8) e^ {0.196x} \\ \text{ } \\ \int {d} \big( y\cdot e^ {0.196x} \big)= (9.8) \int{e^{0.196x} {dx} } \\ \text{ } \\ y\cdot e^ {0.196x} =( \frac{9.8}{0.196}) e^{0.196x}+C \\ \implies y =( \frac{9.8}{0.196})+C\cdot e^ {-0.196x} \\ \therefore y =50+C\cdot e^ {-0.196x}$

Now that we have the general solution we proceed to use y(0) = 48 to find the particular solution:

$\\ y(0)=48 =50+C\cdot e^ {-0.196(0)} \\ \implies 48 =50+C \implies C=-2 \\ \therefore y =50-2\cdot e^ {-0.196x}$

In conclusion, the following initial value problem hasthe particular solution $y =50-2\cdot e^ {-0.196x}$.

Reference:

• Thomas, G. B., & Finney, R. L. (1961). Calculus. Addison-Wesley Publishing Company.