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`,

`-\frac{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=(e^(6x^2)(6x^2-1))/(18)+C`

`\implies y^4=\frac1{(e^(6x^2)(6x^2-1))/(18)+C}`

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

`1=\frac1{(5e^6)/(18)+C}\implies C=(18-5e^6)/(18)`

so the particular solution is

`y^4=\frac1{(e^(6x^2)(6x^2-1))/(18)+(18-5e^6)/(18)}`

`\implies\boxed{y=\sqrt[4]{\frac1{(e^(6x^2)(6x^2-1))/(18)+(18-5e^6)/(18)}}}`

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