4.15
[tex]y'+\dfrac 2xy=x^3[/tex]
is a linear ODE; multiply both sides by the integrating factor [tex]x^2[/tex]:
[tex]x^2y'+2xy=x^5[/tex]
Now the left side can be condensed as the derivative of a product:
[tex](x^2y)'=x^5[/tex]
Integrate both sides and solve for [tex]y[/tex] to get
[tex]x^2y=\dfrac{x^6}6+C\implies y=\dfrac{x^4}6+\dfrac C{x^2}[/tex]
Given that [tex]y(1)=-\frac56[/tex], we find
[tex]-\dfrac56=\dfrac16+C\implies C=-1[/tex]
and so the particular solution is
[tex]\boxed{y(x)=\dfrac{x^4}6-\dfrac1{x^2}}[/tex]
5.15
[tex]2y^2\,\mathrm dx+(x+e^{1/y})\,\mathrm dy=0[/tex]
You may be tempted to write this as an ODE in [tex]y(x)[/tex], but the ODE in [tex]x(y)[/tex] is much easier to solve, since it's linear. Solve for [tex]\frac{\mathrm dx}{\mathrm dy}=x'[/tex] and rearrange the terms:
[tex]\dfrac{\mathrm dx}{\mathrm dy}=-\dfrac{x+e^{1/y}}{2y^2}\implies x'+\dfrac x{2y^2}=-\dfrac{e^{1/y}}{2y^2}[/tex]
Multiply both sides by the integrating factor [tex]e^{-1/(2y)}[/tex], then solve for [tex]x[/tex]:
[tex]e^{-1/(2y)}x'+\dfrac{e^{-1/(2y)}}{2y^2}x=-\dfrac{e^{1/(2y)}}{2y^2}[/tex]
[tex]\left(e^{-1/(2y)}x\right)'=-\dfrac{e^{1/(2y)}}{2y^2}[/tex]
[tex]e^{-1/(2y)}x=e^{1/(2y)}+C[/tex]
[tex]x=e^{1/y}+Ce^{1/(2y)}[/tex]
Given that [tex]y=1[/tex] when [tex]x=e[/tex], we have
[tex]e=e+Ce^{1/2}\implies C=0[/tex]
so the particular solution is
[tex]\boxed{x(y)=e^{1/y}}[/tex]
or, by solving for [tex]y[/tex],
[tex]\boxed{y(x)=\dfrac1{\ln x}}[/tex]
6.15
[tex]y'-y=2xy^2[/tex]
Dividing through both sides by [tex]y^2[/tex] lets us write the equation in Bernoulli form:
[tex]y^{-2}y'-y^{-1}=2x[/tex]
Substitute [tex]v=y^{-1}[/tex], so that [tex]v'=-y^{-2}y'[/tex]. Then we get an ODE that is linear in [tex]v[/tex]:
[tex]-v'-v=2x\implies v'+v=-2x[/tex]
Multiply both sides by the integrating factor [tex]e^x[/tex]:
[tex]e^xv'+e^xv=(e^xv)'=-2xe^x[/tex]
Integrate both sides and solve for [tex]v[/tex], then solve for [tex]y[/tex]:
[tex]e^xv=-2e^x(x-1)+C[/tex]
[tex]v=-2(x-1)+Ce^{-x}[/tex]
[tex]\dfrac1y=-2(x-1)+Ce^{-x}[/tex]
Given that [tex]y(0)=\frac12[/tex], we find
[tex]2=2+C\implies C=0[/tex]
so the particular solution is
[tex]\dfrac1y=-2(x-1)=2-2x\implies\boxed{y(x)=\dfrac1{2-2x}}[/tex]
