Solve y' + 2xy = 2x^3, y(0) = 1

asked Mar 15, 2018 124k views

1 vote

8.2k points

Please log in or register to add a comment.

6 votes

Multiply both sides by
e^(x^2)

, then you get

e^(x^2)+2xe^(x^2)y=2x^3e^(x^2)

$\left(e^(x^2)y\right)'=2x^3e^(x^2)$

$e^(x^2)y=\displaystyle2\int x^3e^(x^2)\,\mathrm dx$

e^(x^2)y=e^(x^2)(x^2-1)+C

Given that
y(0)=1

, you have

$1=-1+Ce^0\implies C=2$

so that the particular solution is

y=x^2-1+2e^(-x^2)

Sakshi Singla answered Mar 19, 2018

8.1k points

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers