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An equation in the form y′+p(x)y=q(x)yny′+p(x)y=q(x)yn with n≠0,1n≠0,1 is called a Bernoulli equation and it can be solved using the substitution v=y1−nv=y1−n which transforms the Bernoulli equation into the following first order linear equation for vv:

v′+(1−n)p(x)v=(1−n)q(x)v′+(1−n)p(x)v=(1−n)q(x)

Given the Bernoulli equation

y′+4xy=12xy34 (∗)y′+4xy=12xy34 (∗)

we have n=n= so v=v= .We obtain the equation v′+ v′+ v=v= .Solving the resulting first order linear equation for vv we obtain the general solution (with arbitrary constant CC) given by

v=v=

Then transforming back into the variables xx and yy and using the initial condition y(1)=1y(1)=1 to find C=C= .Finally we obtain the explicit solution of the initial value problem as

User Siddhant
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The given equation is a Bernoulli equation, which can be solved using the substitution v = y⁽¹⁻ⁿ⁾. By substituting v into the equation, we obtain the first order linear equation v' + (1-n)p(x)v = (1-n)q(x).


In this case, the given Bernoulli equation is y' + 4xy = 12xy³. We have n = 3/4, so v = y^(1 - 3/4) = y^(1/4).

Substituting v into the equation, we get v' + (1 - 3/4)(4x)v = (1 - 3/4)(12x). Simplifying this equation, we have v' + (1/4)(4x)v = (1/4)(12x).

Now, we solve this resulting first order linear equation for v. Integrating both sides, we get v = Ce^(-x/4) + 3x. Here, C is an arbitrary constant.

To find the explicit solution of the initial value problem, we substitute v back into the equation v = y^(1/4), which gives y = (Ce^(-x/4) + 3x)^4.

To find the value of C, we use the initial condition y(1) = 1. Substituting x = 1 and y = 1 into the equation, we solve for C.

Finally, the explicit solution of the initial value problem is y = (Ce^(-x/4) + 3x)^4, where C is the value obtained from the initial condition.

This is the step-by-step process to solve the given Bernoulli equation and find the explicit solution of the initial value problem.

User Fchauvel
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