Question

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= .

Answer 1

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




then the ODE will be exact if
`M_y=N_x`. We have


`M_y=4`

`N_x=16y^3e^(-4x)`

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


`\mu M\,\mathrm dx+\mu N\,\mathrm dy=0`

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


`(\mu M)_y=(\mu N)_x\implies\mu_yM+\mu M_y=\mu_xN+\mu N_x`

`\implies\mu_yM-\mu_xN=\mu(N_x-M_y)`

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


`\mu_x N=\mu(M_y-N_x)\implies\frac{\mathrm d\mu}\mu=\frac{M_y-N_x}N\,\mathrm dx`

So in our case we might consider using


`\frac{\mathrm d\mu}\mu=(4-16y^3e^(-4x))/(1-4y^3e^(-4x))\,\mathrm dx=4\,\mathrm dx`

`\implies\displaystyle\int\frac{\mathrm d\mu}\mu=4\int\mathrm dx`

`\implies\ln|\mu|=4x`

`\implies\mu=e^(4x)`

Our new ODE is guaranteed to be exact:


`(4ye^(4x)+5x^4)\,\mathrm dx+(e^(4x)-4y^3)\,\mathrm dy=0`

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


`(\mathrm df)/(\mathrm dx)=(\partial f)/(\partial x)(\mathrm dx)/(\mathrm dx)+(\partial f)/(\partial y)(\mathrm dy)/(\mathrm dx)=0`

`\implies(\partial f)/(\partial x)+(\partial f)/(\partial y)(\mathrm dy)/(\mathrm dx)=0`

`\implies(\partial f)/(\partial x)\,\mathrm dx+(\partial df)/(\partial y)\mathrm dy=0`

Now,


`(\partial f)/(\partial x)=\mu M=4ye^(4x)+5x^4`

`\implies f=ye^(4x)+x^5+g(y)`

Differentiating with respect to
`y` gives


`(\partial f)/(\partial y)=\mu N`

`\implies e^(4x)+(\mathrm dg)/(\mathrm dy)=e^(4x)-4y^3`

`\implies(\mathrm dg)/(\mathrm dy)=-4y^3`

`\implies g(y)=-y^4+C`

So the general solution is


`f(x,y)=ye^(4x)+x^5-y^4+C=C`

`\implies f(x,y)=ye^(4x)+x^5-y^4=C`

Given that
`y(0)=1`, we get


`f(0,1)=1-1^4=0=C`

so the particular solution is just


`ye^(4x)+x^5-y^4=0`