How to you find the general solution of #4yy'-3e^x=0#?

Question

Aurora Alvarado · Accepted Answer

# y = sqrt(3/2e^x + C) #

Changing the notation from Lagrange's notation to Leibniz's notation we have:

# 4ydy/dx-3e^x = 0 # # :. 4ydy/dx = 3e^x #

This is a First Order sepaable DE, and "seperating the variables" give is

# :. int 4y dy = int3e^x dx #

Integrating gives us:

# 2y^2 = 3e^x + C_1 # # :. y^2 = 3/2e^x + C # (by writing #C_1=1/2C#) # :. y = sqrt(3/2e^x + C) #

Charles Arocha · Answer

undefined To find the general solution of the differential equation $4yy' - 3e^x = 0$, you can use the method of separation of variables. Rearrange the equation to isolate $y$ and $y'$ terms. Then integrate both sides with respect to $x$ to solve for $y$. The general solution will contain an arbitrary constant.

Starting with the given equation $4yy' - 3e^x = 0$, we rearrange to isolate the $y'$ term:

$$4yy' = 3e^x$$

$$y' = \frac{3e^x}{4y}$$

Now, separate the variables by multiplying both sides by $dx$ and dividing both sides by $y$:

$$\frac{dy}{dx} = \frac{3e^x}{4y}$$

$$4y \, dy = 3e^x \, dx$$

Next, integrate both sides:

$$\int 4y \, dy = \int 3e^x \, dx$$

$$2y^2 = 3e^x + C$$

Finally, solve for $y$:

$$y^2 = \frac{3}{2} e^x + \frac{C}{2}$$

$$y = \pm \sqrt{\frac{3}{2} e^x + \frac{C}{2}}$$

So, the general solution of the differential equation is:

$$y = \pm \sqrt{\frac{3}{2} e^x + \frac{C}{2}}$$

Where $C$ is the constant of integration.