Question

Show complete solution. Use equation function.

Find the particular solution of dy/dx = ycosx/1+2y^2 given y(x) = 1; y (0) = 1

Answer 1

We first separate all the terms to find the equations that have to be integrated to find y(x):

$\frac{dy}{dx}=\frac{y\,cosx}{1+2y^2} \implies \int (\frac{1+2y^2}{y}) dy= \int cosx \, dx$

$\int (\frac{1}{y}+2y) dy= \int cosx \, dx \implies \ln y+y^2=sinx+C$

After that, we use y(0)=1 as the boundary condition to find C and the particular solution:

$\ln y+y^2=sinx+C \implies \ln (1)+(1)^2-sin(0)=C \implies C=1$

In conclusion, the particular solution for the DE is $sinx=\ln y+y^2-1$.

Reference:

• Zill, D. G. (2016). Differential equations with boundary-value problems. Cengage Learning.