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

$\dfrac{u'}{u^2-1}=1$

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

$\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1$

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

$\dfrac12(\ln|u-1|-\ln|u+1|)=t+C$

Solve for $u$:

$\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C$

$\ln\left|1-\dfrac2{u+1}\right|=2t+C$

$1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}$

$\dfrac2{u+1}=1-Ce^{2t}$

$\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}$

$u=\dfrac2{1-Ce^{2t}}-1$

Replace $u$ and solve for $y$:

$t+y=\dfrac2{1-Ce^{2t}}-1$

$y=\dfrac2{1-Ce^{2t}}-1-t$

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

$y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}$

so that the particular solution is

$y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}$