Solved on Feb 27, 2024

Solve the differential equation 3yy=5x3 y y' = 5 x.

STEP 1

Assumptions
1. The given differential equation is 3yy=5x3yy' = 5x.
2. yy' represents the derivative of yy with respect to xx.
3. We are looking for the general solution y(x)y(x) of the differential equation.

STEP 2

Rewrite the differential equation in a separable form, where all terms involving yy are on one side and all terms involving xx are on the other side.
1ydy=531xdx\frac{1}{y} dy = \frac{5}{3} \frac{1}{x} dx

STEP 3

Integrate both sides of the equation with respect to their respective variables.
1ydy=531xdx\int \frac{1}{y} dy = \int \frac{5}{3} \frac{1}{x} dx

STEP 4

Perform the integration on both sides.
lny=53lnx+C\ln |y| = \frac{5}{3} \ln |x| + C
Here, CC is the constant of integration.

STEP 5

Exponentiate both sides to remove the natural logarithm and solve for yy.
elny=e53lnx+Ce^{\ln |y|} = e^{\frac{5}{3} \ln |x| + C}

STEP 6

Simplify the equation using the property elna=ae^{\ln a} = a.
y=eCx53|y| = e^{C} \cdot |x|^{\frac{5}{3}}

STEP 7

Since y|y| can be either yy or y-y, and eCe^{C} is a constant, we can write eCe^{C} as a new constant kk where k>0k > 0. We can also drop the absolute value since for every positive kk, there is a corresponding negative k-k that will work for negative yy.
y=kx53y = kx^{\frac{5}{3}}
Here, kk is a constant that can be any real number.
The general solution of the differential equation 3yy=5x3yy' = 5x is:
y=kx53y = kx^{\frac{5}{3}}

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