# Math Help - IVP

1. ## IVP

Hello,
I was given the problem and to solve the initial value problem

5x4y3dx+3x5y2dy=0y(1)=1

I was told to find the answer in terms of F(x,y)=1 and F(x,y) does not have a constant.

I got to the part where i've integrated both sides and got

5ln(x) + 3ln(y)

I'm just confused on what to do next, i tried solving for the constant which is 0

but have no clue what the question means when its asking for it in terms of F(x,y)=1

both answers i tried were wrong: 1- 5ln(x) + 3ln(y) & 2- 5ln(x) + 3ln(y) + 1

thankss

2. Originally Posted by washerman
Hello,
I was given the problem and to solve the initial value problem

5x4y3dx+3x5y2dy=0y(1)=1

I was told to find the answer in terms of F(x,y)=1 and F(x,y) does not have a constant.

I got to the part where i've integrated both sides and got

5ln(x) + 3ln(y)

I'm just confused on what to do next, i tried solving for the constant which is 0

but have no clue what the question means when its asking for it in terms of F(x,y)=1

both answers i tried were wrong: 1- 5ln(x) + 3ln(y) & 2- 5ln(x) + 3ln(y) + 1

thankss
The ODE is first order linear and separable
$5x^4y^3dx+3x^5y^2dy=0 \iff \frac{dy}{dx}=-\frac{5}{3}\cdot \frac{y}{x} \iff \frac{dy}{y}=-\frac{5}{3}\cdot \frac{dx}{x}$

Integrating both sides gives

$\ln|y|=-\frac{5}{3}\ln|x|+c \iff y=Ax^{-\frac{5}{3}}$

Using the initial condition gives A=1

$y=x^{-\frac{5}{3}}$

To put the solution into implicit form we have

$x^{-\frac{5}{3}}-y=0 \text{ or } x^{-\frac{5}{3}}-y+1=1$