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In this problem we consider an equation in differential form mdx+ndy=0. the equation (4y+(5x^4)e^(?4x))dx+(1?4y^3(e^(?4x)))dy=0 in differential form m˜dx+n˜dy=0 is not exact. indeed, we have m˜y?n˜x= for this exercise we can find an integrating factor which is a function of x alone since m˜y?n˜xn˜= can be considered as a function of x alone. namely we have ?(x)= multiplying the original equation by the integrating factor we obtain a new equation mdx+ndy=0 where m= n= which is exact since my= nx= are equal. this problem is exact. therefore an implicit general solution can be written in the form f(x,y)=c where f(x,y)= finally find the value of the constant c so that the initial condition y(0)=1. c= .

Respuesta :

LammettHash
Taking a wild guess as to what those question marks are supposed to encode... If the ODE is

[tex]\underbrace{(4y+5x^4e^{-4x})_{M(x,y)}\,\mathrm dx+\underbrace{(1-4y^3e^{-4x})}_{N(x,y)}\,\mathrm dy=0[/tex]

then the ODE will be exact if [tex]M_y=N_x[/tex]. We have

[tex]M_y=4[/tex]
[tex]N_x=16y^3e^{-4x}[/tex]

and so indeed the equation is not exact. So we look for an integrating factor [tex]\mu(x,y)[/tex] such that

[tex]\mu M\,\mathrm dx+\mu N\,\mathrm dy=0[/tex]

is exact. In order for this to occur, we require

[tex](\mu M)_y=(\mu N)_x\implies\mu_yM+\mu M_y=\mu_xN+\mu N_x[/tex]
[tex]\implies\mu_yM-\mu_xN=\mu(N_x-M_y)[/tex]

Now if [tex]\mu[/tex] is a function of either [tex]x[/tex] or [tex]y[/tex] alone, then this PDE reduces to an ODE in either variable. Let's assume the first case, so that [tex]\mu_y=0[/tex]. Then

[tex]\mu_x N=\mu(M_y-N_x)\implies\dfrac{\mathrm d\mu}\mu=\dfrac{M_y-N_x}N\,\mathrm dx[/tex]

So in our case we might consider using

[tex]\dfrac{\mathrm d\mu}\mu=\dfrac{4-16y^3e^{-4x}}{1-4y^3e^{-4x}}\,\mathrm dx=4\,\mathrm dx[/tex]
[tex]\implies\displaystyle\int\frac{\mathrm d\mu}\mu=4\int\mathrm dx[/tex]
[tex]\implies\ln|\mu|=4x[/tex]
[tex]\implies\mu=e^{4x}[/tex]

Our new ODE is guaranteed to be exact:

[tex](4ye^{4x}+5x^4)\,\mathrm dx+(e^{4x}-4y^3)\,\mathrm dy=0[/tex]

so we can now look for our solution [tex]f(x,y)=C[/tex]. By the chain rule, differentiating with respect to [tex]x[/tex] yields

[tex]\dfrac{\mathrm df}{\mathrm dx}=\dfrac{\partial f}{\partial x}\dfrac{\mathrm dx}{\mathrm dx}+\dfrac{\partial f}{\partial y}\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]
[tex]\implies\dfrac{\partial f}{\partial x}+\dfrac{\partial f}{\partial y}\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]
[tex]\implies\dfrac{\partial f}{\partial x}\,\mathrm dx+\dfrac{\partial df}{\partial y}\mathrm dy=0[/tex]

Now,

[tex]\dfrac{\partial f}{\partial x}=\mu M=4ye^{4x}+5x^4[/tex]
[tex]\implies f=ye^{4x}+x^5+g(y)[/tex]

Differentiating with respect to [tex]y[/tex] gives

[tex]\dfrac{\partial f}{\partial y}=\mu N[/tex]
[tex]\implies e^{4x}+\dfrac{\mathrm dg}{\mathrm dy}=e^{4x}-4y^3[/tex]
[tex]\implies\dfrac{\mathrm dg}{\mathrm dy}=-4y^3[/tex]
[tex]\implies g(y)=-y^4+C[/tex]

So the general solution is

[tex]f(x,y)=ye^{4x}+x^5-y^4+C=C[/tex]
[tex]\implies f(x,y)=ye^{4x}+x^5-y^4=C[/tex]

Given that [tex]y(0)=1[/tex], we get

[tex]f(0,1)=1-1^4=0=C[/tex]

so the particular solution is just

[tex]ye^{4x}+x^5-y^4=0[/tex]

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