A Bernoulli Equation

Calculus Level 1

$\large y'+y=e^xy^3$

If a function $y$ satisfies the differential equation above, and $y(0)=\dfrac{1}{\sqrt{3}}$ , what is $y(1)$ ?

\frac{1}{\sqrt{e^2+e}}

\frac{1}{\sqrt{e^2-2e}}

\frac{1}{\sqrt{e^2+2e}}

\frac{1}{\sqrt{e^2-e}}

1 solution

Samir Khan
Jun 13, 2016

Dividing through by $y^3$ , we have $y'y^{-3}+y^{-2}=e^x.$ Now, if $v=y^{-2}$ , then $v'=-2y^{-3} y'$ , so our equation is $-\frac{1}{2} v'+v=e^x\implies v'-2v=-2e^x.$ Multiplying through by the integrating factor $e^{-2x}$ , we find $v'e^{-2x}-2e^{-2x}v=-2e^{-x}\implies \left[ve^{-2x}\right]'=-2e^{-x}\implies ve^{-2x}=2e^{-x}+C\implies v=Ce^{2x}+2e^x.$ This means that $y^2=\frac{1}{Ce^{2x}+2e^x},$ , and the initial condition tells us that $C=1$ and to take the positive branch of the square root. Thus, $y(x)=\frac{1}{\sqrt{e^{2x}+2e^x}}\implies y(1)=\frac{1}{\sqrt{e^2+2e}}.$

How did you find the integrating factor to be e^-2x?

Aldahir Pereira - 2 years ago

The main aim in the solution is to create an expression that resembles d(xy) form (using the product rule) in the left hand side and then integrating both the sides of the resulting equation to get the answer.

Smriti Prasad - 1 year, 10 months ago

after multiplying I.F. how did the 2nd value (-2(e^(-2x))*v) vanish??

Fazla Alvi - 1 year, 6 months ago

A Bernoulli Equation

1 solution

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