Question

2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One common change of variable technique is as follows. Consider a differential equation of the form y′=f(αt+βy+γ), where α,β, and γ are constants. Use the change of variable z=αt+βy+γ to rewrite the differential equation as a separable equation of the form z′=g(z). Solve the initial value problem y′=(t+y)2−1, y(3)=4.

Answer 1

`y'=(t+y)^2-1`

Substitute
`u=t+y`, so that
`u'=y'`, and

`u'=u^2-1`

which is separable as

`(u')/(u^2-1)=1`

Integrate both sides with respect to
`t`. For the integral on the left, first split into partial fractions:

`\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1`

`\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt`

`\frac12(\ln|u-1|-\ln|u+1|)=t+C`

Solve for
`u`:

`\frac12\ln\left|(u-1)/(u+1)\right|=t+C`

`\ln\left|1-\frac2{u+1}\right|=2t+C`

`1-\frac2{u+1}=e^(2t+C)=Ce^(2t)`

`\frac2{u+1}=1-Ce^(2t)`

`\frac{u+1}2=\frac1{1-Ce^(2t)}`

`u=\frac2{1-Ce^(2t)}-1`

Replace
`u` and solve for
`y`:

`t+y=\frac2{1-Ce^(2t)}-1`

`y=\frac2{1-Ce^(2t)}-1-t`

Now use the given initial condition to solve for
`C`:

`y(3)=4\implies4=\frac2{1-Ce^6}-1-3\implies C=\frac3{4e^6}`

so that the particular solution is

`y=\frac2{1-\frac34e^(2t-6)}-1-t=\boxed{\frac8{4-3e^(2t-6)}-1-t}`