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In Problems \(15\) and \(16\) interpret each statement as a differential equation.

On the graph of \(y = \phi (x)\) the slope of the tangent line at a point \(P(x,y)\) is the square of the distance from \(P(x,y)\) to the origin.

Short Answer

Expert verified

The differential equation is \(\frac{{dy}}{{dx}} = {x^2} + {y^2}\).

Step by step solution

01

Define a derivative of the function.

The derivative of a function of a real variable in mathematics describes the sensitivity of the function value (output value) to changes in its argument (input value).

Calculus uses derivatives as a fundamental tool. When a derivative of a single-variable function exists at a given input value, it is the slope of the tangent line to the function's graph at that point.

02

Determine the differential equation for the height.

Let the slope of the tangent line of the graph of\(y = \phi (x)\)at the point\(P(x,y)\)is\(\frac{{dy}}{{dx}}\).

Using the Pythagoras theorem, the square of the distance of\(P(x,y)\)to the origin is\({x^2} + {y^2}\).

Hence, the statement describes the differential equation,

\(\frac{{dy}}{{dx}} = {x^2} + {y^2}\).

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Most popular questions from this chapter

In Problems 21–24 verify that the indicated family of functions is a solution of the given differential equation. Assume an appropriate interval I of definition for each solution.

\(\frac{{dP}}{{dt}} = P(1 - P);\;P = \frac{{{c_1}{e^t}}}{{1 + {c_1}{e^t}}}\)

(a) Verify that the one-parameter family \({y^2} - 2y = {x^2} - x + c\) is an implicit solution of the differential equation \((2y - 2)y' = 2x - 1\).

(b) Find a member of the one-parameter family in part (a) that satisfies the initial condition \(y(0) = 1\).

(c) Use your result in part (b) to and an explicit function \(y = \phi (x)\) that satisfies \(y(0) = 1\). Give the domain of the function \(\phi \). Is \(y = \phi (x)\) a solution of the initial-value problem? If so, give its interval \(I\) of definition; if not, explain.

In Problems \(15 - 18\) verify that the indicated functionis an explicit solution of the given first-order differential equation. Proceed as in Example \(6\), by considering \(\phi \) simply as a function and give its domain. Then by considering \(\phi \) as a solution of the differential equation, give at least one interval \(I\) of definition.

\(y' = 25 + {y^2};y = 5tan5x\)

In Problems 25–28 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\]

In Problems 27–30 use (12) of Section 1.1 to verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval I of definition of each solution.

\({x^2}y'' + \left( {{x^2} - x} \right)y' + (1 - x)y = 0;\;\;\;y = x\int_1^x {\frac{{{e^{ - t}}}}{t}} dt\)

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