Solution of the differential equation $\ln \bigg(\frac{dy}{dx} \bigg)=e^{ax+by}$

Question

Solve the given differential equation:

$$\ln \bigg(\frac{dy}{dx} \bigg)=e^{ax+by}$$

Could someone give me guidance to proceed in this question. I am not able to deal with $\frac{dy}{dx}=e^{e^{ax+by}}$

Answer 1

Let $z= ax + by$

The $\frac{dz}{dx} = a + b \frac{dy}{dx}$

So we have that $\frac{1}{b} \frac{dz}{dx} - \frac{a}{b} = \frac{dy}{dx}$ thus:

$$\frac{1}{b} \frac{dz}{dx} - \frac{a}{b} = e^{e^z}$$

Balancing both sides we have

$$ \frac{1}{ a+b e^{e^z} } dz = 1 dx$$

To make the left side nicer consider the L substitution $L = e^{e^z}, dL = e^z e^{e^z} dz \rightarrow dz = \frac{dL}{L \ln(L)}$

This yields

$$ \int \frac{1}{\ln(L)}\frac{1}{aL + bL^2} dL = x+C$$

As a newer nicer looking problem to solve.

Next we consider that $(\sqrt{b}L +q)^2 =bL^2 + 2q\sqrt{b} +q^2$ So it follows that we have

$$ \int \frac{1}{\ln(L)}\frac{1}{(\sqrt{b}L + \frac{a}{2\sqrt{b}})^2 - \frac{a^2}{4b}} dL = x+C$$

And I don't think this can be done in elementary means