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Chapter 8: Q78SE (page 359)

In Problems 25鈥�28 use (12) to verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval I of definition of each solution.

\(2x\frac{{dy}}{{dx}} - y = 2xcosx;y = \sqrt x \int_4^x {\frac{{cost}}{{\sqrt t }}} dt\)

Short Answer

Expert verified

The indicated function is a solution of the differential function.

Step by step solution

01

Simplify the given differential equation.

Let the given differential equation be \(y = \sqrt x \int_4^x {\frac{{cost}}{{\sqrt t }}} dt\).

Multiply each side of the equation by \({x^{ - \frac{1}{2}}}\).

\(\begin{aligned}{c}y{x^{ - \frac{1}{2}}} &= {x^{ - \frac{1}{2}}}{x^{\frac{1}{2}}}\int_4^x {\frac{{cost}}{{\sqrt t }}} \;dt\\y{x^{ - \frac{1}{2}}} &= \int_4^x {\frac{{cost}}{{\sqrt t }}} \;dt\end{aligned}\)

02

Determine the solution of the indicated function.

Take differential on both sides of the equation.

Multiply \(2{x^{\frac{3}{2}}}\) on both sides of the equation.

\(\begin{aligned}{c}2x\frac{{dy}}{{\;dx}} - y &= 2{x^{\frac{3}{2}}}{x^{ - \frac{1}{2}}}cosx\\2x\frac{{dy}}{{\;dx}} - y &= 2xcosx\end{aligned}\)

Hence, the indicated function is a solution of the differential function.

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