Question

Please help with differential equations

Answer 1

4.15

`y'+\frac 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=\frac{x^6}6+C\implies y=\frac{x^4}6+\frac C{x^2}`

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

`-\frac56=\frac16+C\implies C=-1`

and so the particular solution is

`\boxed{y(x)=\frac{x^4}6-\frac1{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
`(\mathrm dx)/(\mathrm dy)=x'` and rearrange the terms:

`(\mathrm dx)/(\mathrm dy)=-(x+e^(1/y))/(2y^2)\implies x'+\frac x{2y^2}=-(e^(1/y))/(2y^2)`

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

`e^(-1/(2y))x'+(e^(-1/(2y)))/(2y^2)x=-(e^(1/(2y)))/(2y^2)`

`\left(e^(-1/(2y))x\right)'=-(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)=\frac1{\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)`

`\frac1y=-2(x-1)+Ce^(-x)`

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

`2=2+C\implies C=0`

so the particular solution is

`\frac1y=-2(x-1)=2-2x\implies\boxed{y(x)=\frac1{2-2x}}`