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Find the general solution of the Bernoulli differential equation: y' + 4y = xy^4. Use lower case c for the constant in answer.

User Krisja
by
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1 Answer

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Answer with Step-by-step explanation:

The given equation is


(dy)/(dx)+4y=xy^4

Dividing throughout by
y^4 we get


(dy)/(y^4dx)+(4)/(y^3)=x\\\\y^(-4)(dy)/(dx)+4y^(-3)=x\\\\

Substituting
t=y^(-3) in the above equation we get


dt=-3y^(-4)\cdot (dy)/(dx)\\\\y^(-4)\cdot {dy}=(-dt)/(3)

Thus we get


(-dt)/(dx)+12t=3x

Which is a linear differential equation of the form


t'+p(x)t=q(x)dx}[/tex]

whose solution is given by


te^(\int p(x)dx)=\int (e^(\int p(x))q(x)dx

Solving we get


te^(12x)=\int 3e^(12x)* xdx\\\\te^(12x)=3x\cdot (e^(12x))/(12)-\int 3* (e^(12x))/(12)dx\\\\te^(12x)=(xe^(12x))/(4)-(e^(12x))/(48)\\\\t=(x)/(4)-(1)/(48)\\\\(1)/(y^3)=(x)/(4)-(1)/(48)+c

User Mlg
by
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