Question

Please help with differential equations

Answer 1

4.15

$y'+\dfrac 2xy=x^3$

is a linear ODE; multiply both sides by the integrating factor $x^2$:

$x^2y'+2xy=x^5$

Now the left side can be condensed as the derivative of a product:

$(x^2y)'=x^5$

Integrate both sides and solve for $y$ to get

$x^2y=\dfrac{x^6}6+C\implies y=\dfrac{x^4}6+\dfrac C{x^2}$

Given that $y(1)=-\frac56$, we find

$-\dfrac56=\dfrac16+C\implies C=-1$

and so the particular solution is

$\boxed{y(x)=\dfrac{x^4}6-\dfrac1{x^2}}$

5.15

$2y^2\,\mathrm dx+(x+e^{1/y})\,\mathrm dy=0$

You may be tempted to write this as an ODE in $y(x)$, but the ODE in $x(y)$ is much easier to solve, since it's linear. Solve for $\frac{\mathrm dx}{\mathrm dy}=x'$ and rearrange the terms:

$\dfrac{\mathrm dx}{\mathrm dy}=-\dfrac{x+e^{1/y}}{2y^2}\implies x'+\dfrac x{2y^2}=-\dfrac{e^{1/y}}{2y^2}$

Multiply both sides by the integrating factor $e^{-1/(2y)}$, then solve for $x$:

$e^{-1/(2y)}x'+\dfrac{e^{-1/(2y)}}{2y^2}x=-\dfrac{e^{1/(2y)}}{2y^2}$

$\left(e^{-1/(2y)}x\right)'=-\dfrac{e^{1/(2y)}}{2y^2}$

$e^{-1/(2y)}x=e^{1/(2y)}+C$

$x=e^{1/y}+Ce^{1/(2y)}$

Given that $y=1$ when $x=e$, we have

$e=e+Ce^{1/2}\implies C=0$

so the particular solution is

$\boxed{x(y)=e^{1/y}}$

or, by solving for $y$,

$\boxed{y(x)=\dfrac1{\ln x}}$

6.15

$y'-y=2xy^2$

Dividing through both sides by $y^2$ lets us write the equation in Bernoulli form:

$y^{-2}y'-y^{-1}=2x$

Substitute $v=y^{-1}$, so that $v'=-y^{-2}y'$. Then we get an ODE that is linear in $v$:

$-v'-v=2x\implies v'+v=-2x$

Multiply both sides by the integrating factor $e^x$:

$e^xv'+e^xv=(e^xv)'=-2xe^x$

Integrate both sides and solve for $v$, then solve for $y$:

$e^xv=-2e^x(x-1)+C$

$v=-2(x-1)+Ce^{-x}$

$\dfrac1y=-2(x-1)+Ce^{-x}$

Given that $y(0)=\frac12$, we find

$2=2+C\implies C=0$

so the particular solution is

$\dfrac1y=-2(x-1)=2-2x\implies\boxed{y(x)=\dfrac1{2-2x}}$