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Chapter 6: Problem 27

Solve the differential equation. $$ \frac{d y}{d t}=\frac{t^{2}}{\sqrt{2+3 t}} $$

Short Answer

Expert verified
The solution to the differential equation \(\frac{dy}{dt} = \frac{t^2}{\sqrt{2+3t}}\) is \(y = \frac{1}{18}(2+3t)^{\frac{3}{2}} - \frac{4}{3}(2+3t)^{\frac{1}{2}} + \frac{8}{3\sqrt{2+3t}} + C\).

Step by step solution

01

Integrate both sides of the equation

The given equation is \(\frac{dy}{dt} = \frac{t^2}{\sqrt{2+3t}}\). Let's start by taking an integral on both sides. On the left side this is a simple integral, resulting in \(y\). On the right side, using the substitution \(x=2+3t\), we end up with \(\int \frac{t^2}{\sqrt{2+3t}} dt = y + C\), where \(C\) is the constant of integration.
02

Perform substitution

Let us perform the substitution. Let \(x = 2 + 3t\), then \(\frac{dx}{dt} = 3\) ==> \(dt = \frac{dx}{3}\). Therefore, the integral can be rewritten as \(\int \frac{(x-2)^2}{3\sqrt{x}} dx\). This simplifies the calculation.
03

Divide through by 3 before taking the integral

The revised equation after division by 3 is \(\int \frac{(x-2)^2}{\sqrt{x}} dx\). Apply the power rule for integration along with sum and difference rule and solve the integral.
04

Solve and Simplify

The integral, when solved, provides \(y = \frac{1}{18}x^{\frac{3}{2}} - \frac{4}{3}x^{\frac{1}{2}} + \frac{8}{3\sqrt{x}} + C\). Substitute \(x = 2+3t\) back into the equation to get the final solution.

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