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Andre45 [30]
3 years ago
13

Please I need help with differential equation. Thank you

Mathematics
1 answer:
Inga [223]3 years ago
3 0

1. I suppose the ODE is supposed to be

\mathrm dt\dfrac{y+y^{1/2}}{1-t}=\mathrm dy(t+1)

Solving for \dfrac{\mathrm dy}{\mathrm dt} gives

\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{y+y^{1/2}}{1-t^2}

which is undefined when t=\pm1. The interval of validity depends on what your initial value is. In this case, it's t=-\dfrac12, so the largest interval on which a solution can exist is -1\le t\le1.

2. Separating the variables gives

\dfrac{\mathrm dy}{y+y^{1/2}}=\dfrac{\mathrm dt}{1-t^2}

Integrate both sides. On the left, we have

\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\int\frac{\mathrm dz}{z+1}

where we substituted z=y^{1/2} - or z^2=y - and 2z\,\mathrm dz=\mathrm dy - or \mathrm dz=\dfrac{\mathrm dy}{2y^{1/2}}.

\displaystyle\int\frac{\mathrm dy}{y^{1/2}(y^{1/2}+1)}=2\ln|z+1|=2\ln(y^{1/2}+1)

On the right, we have

\dfrac1{1-t^2}=\dfrac12\left(\dfrac1{1-t}+\dfrac1{1+t}\right)

\displaystyle\int\frac{\mathrm dt}{1-t^2}=\dfrac12(\ln|1-t|+\ln|1+t|)+C=\ln(1-t^2)^{1/2}+C

So

2\ln(y^{1/2}+1)=\ln(1-t^2)^{1/2}+C

\ln(y^{1/2}+1)=\dfrac12\ln(1-t^2)^{1/2}+C

y^{1/2}+1=e^{\ln(1-t^2)^{1/4}+C}

y^{1/2}=C(1-t^2)^{1/4}-1

I'll leave the solution in this form for now to make solving for C easier. Given that y\left(-\dfrac12\right)=1, we get

1^{1/2}=C\left(1-\left(-\dfrac12\right)^2\right))^{1/4}-1

2=C\left(\dfrac54\right)^{1/4}

C=2\left(\dfrac45\right)^{1/4}

and so our solution is

\boxed{y(t)=\left(2\left(\dfrac45-\dfrac45t^2\right)^{1/4}-1\right)^2}

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