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1 day ago

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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.

1 answer:

Svet_ta [4.3K]1 day ago

3 0

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

Substituting $u=t+y$ , yields $u'=y'$ , and

$u'=u^2-1$

which can be separated as

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

Integrate both sides concerning $t$ . For the left integral, split into partial fractions:

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

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

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

Resolve for $u$ :

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

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

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

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

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

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

Substitute $u$ and determine $y$ :

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

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

Utilize the given initial condition to find $C$ :

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

concluding with the particular solution:

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

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