Question

Find the particular solution of each of the differential equation expressing y explicitly in terms of x.

c) (x+4)dy/dx + 3 = y y=13 when x= 1

d) e^ydy/dx + sinx = 0 y=0 when x = π/2

Answer 1

c) We first have to solve the general equation:

$(x+4)\cfrac{dy}{dx} + 3 = y \\ (x+4)\cfrac{dy}{dx}=y-3 \implies \cfrac{dy}{dx}=\cfrac{y-3}{x+4}$

Once we separate the terms we proceed to solve the equation:

$\int \frac{dy}{y-3}=\int \frac{dx}{x+4} \\\to \ln(y-3)=\ln(x+4)+\ln C \\\implies y_{general}=C(x+4)+3;$

We proceed to find the particular solution and

$\\ \text{then we use }y(1)=13: \\13=C(1+4)+3 \to C=\frac{10}{5}=2 \\ \implies y_{particular}=2(x+4)+3 \\ \implies y_{particular}=2x+11$

d) For the next equation we proceed to separate the terms:

$e^y\cfrac{dy}{dx}+\sin (x) =0 \to e^y{dy}=-\sin (x)dx \\ \text{ } \\ \int e^y{dy}=- \int \sin (x)dx \\ \to e^y= \cos (x)+C \\ \implies y=\ln[\cos (x)+C]$

Once we have the general solution we proceed

to find the particular solution so

$\\ \text{then we use }y(π/2)=0: \\ e^0= \cos (π/2)+C \\ \to C=e^0-\cos (π/2)=1-0 \to C=1 \\ \implies y_{particular}=\ln[\cos (x)+1]$

In conclusion, the particular solutions for the differential equations are c) y=2x+11, and d) y=Ln[cos(x)+1].Reference:

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