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Chapter 8: Q77SE (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.

\(x\frac{{dy}}{{dx}} - 3xy = 1;y = {e^{3x}}\int_1^x {\frac{{{e^{ - 3t}}}}{t}} dt\)

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 \(y = {e^{3x}}\int_1^x {\frac{{{e^{ - 3t}}}}{t}} dt\).

Multiply each side of the equation by \({e^{ - 3x}}\).

\(\begin{array}{l}y{e^{ - 3x}} = {e^{3x}}{e^{ - 3x}}\int_1^x {\frac{{{e^{ - 3t}}}}{t}} \;dt\\y{e^{ - 3x}} = \int_1^x {\frac{{{e^{ - 3t}}}}{t}} \;dt\end{array}\)

02

Determine the solution of the indicated function.

Take differential on both sides of the equation.

Multiply \(x\) on both sides of the equation.

\(x{e^{ - 3x}}\frac{{dy}}{{\;dx}} - 3yx{e^{ - 3x}} = {e^{ - 3x}}\)

Divide\({e^{ - 3x}}\)on both sides of the equation.

\(x\frac{{dy}}{{\;dx}} - 3yx = 1\)

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

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