Question

(1 point) A Bernoulli differential equation is one of the form dydx+P(x)y=Q(x)yn. Observe that, if n=0 or 1, the Bernoulli equation is linear. For other values of n, the substitution u=y1−n transforms the Bernoulli equation into the linear equation dudx+(1−n)P(x)u=(1−n)Q(x). Use an appropriate substitution to solve the equation y′−3xy=y5x3, and find the solution that satisfies y(1)=1.

Answer 1

$y'-3xy=y^5x^3$

Divide both sides by $y^5$:

$y^{-5}y'-3xy^{-4}=x^3$

Substitute $u(x)=y(x)^{-4}$, so that $u'=-4y^{-5}y'$. Then we get a new ODE that is linear in $u$,

$-\dfrac{u'}4-3xu=x^3\implies u'+12xu=4x^3$

Multiply both sides by $e^{6x^2}$:

$e^{6x^2}u'+12xe^{6x^2}u=4x^3e^{6x^2}$

Notice that $\left(e^{6x^2}u(x)\right)'=e^{6x^2}u'(x)+12xe^{6x^2}u(x)$. Integrating both sides with respect to $x$ gives

$\displaystyle\int\left(e^{6x^2}u\right)'\,\mathrm dx=e^{6x^2}u=\int 4x^3e^{6x^2}\,\mathrm dx$

$\implies u=\dfrac{e^{6x^2}(6x^2-1)}{18}+C$

$\implies y^4=\dfrac1{\frac{e^{6x^2}(6x^2-1)}{18}+C}$

Given that $y(1)=1$, we find

$1=\dfrac1{\frac{5e^6}{18}+C}\implies C=\dfrac{18-5e^6}{18}$

so the particular solution is

$y^4=\dfrac1{\frac{e^{6x^2}(6x^2-1)}{18}+\frac{18-5e^6}{18}}$

$\implies\boxed{y=\sqrt[4]{\dfrac1{\frac{e^{6x^2}(6x^2-1)}{18}+\frac{18-5e^6}{18}}}}$

where we take the positive root because the initial condition tells us to expect $y>0$ when $x=1$.