Question

1. Use separation of variables to find the solution to the differential equation subject to the given initial condition.
dy/dx = 5y/x, y = 4 while x = 1
2. Find the solution to the differential equation, subject to the given initial condition.
4(du/dt) = u^2, u(0) = 7

Answer 1

Answer:

Step-by-step explanation:

Given the differential equation dy/dx = 5y/x subject to the condition y = 4 and x = 1. Using the variable separable method of solving differential equation, we will have;

dy/dx = 5y/x

Separate the variables

dy/5y = dx/x

Integrate both sides of the expression

$\frac{1}{5}\int\limits \frac{1}{y} \, dy = \int\limits \frac{dx}{x} \\ \\\frac{1}{5}lny = lnx + C\\\\lny = 5lnx+5C$

using the initial condition y = 4 while x = 1

ln4 = 5ln1 + 5C

ln4 = 0+5C

C = ln4/5

Substituting the value of C back into the expression;

$lny = 5 lnx+5(ln4/5)\\lny = 5lnx+ln4\\lny = lnx^5+ln4\\lny = ln(4x^5)\\y = 4x^5$

Hence the solution to the differential equation is y = 4x⁵

b) Given 4(du/dt) = u²

du/dt = u²/4

du/ u² = dt/4

u⁻²du = 1/4 dt

integrate both sides of the equation

$\int\limit {u^{-2}} \, du = \int\limits\frac{1}{4} \, dt\\\\\frac{u^{-1}}{-1} = \frac{t}{4} + C\\\\\frac{-1}{u} = \frac{t}{4} + C$

Imputing the initial condition u(0) = 7 i.e when t = 0, u = 7

$\frac{-1}{7} = \frac{0}{4} + C\\\\\frac{-1}{7} = C$

$\frac{-1}{u} = \frac{t}{4} - \frac{1}{7}$

Hence the solution to the DE is $\frac{-1}{u} = \frac{t}{4} - \frac{1}{7}$