Show that \(y = \frac{1}{x}\int {f(x){\text{d}}x} \) is a solution of the differential equation

\(x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = f(x),{\text{ }}x > 0\).

Hence solve \(x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = {x^{ - \frac{1}{2}}},{\text{ }}x > 0\), given that \(y = 2\) when \(x = 4\).

METHOD 1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - \frac{1}{{{x^2}}}\int {f(x){\text{d}}x + \frac{1}{x}f(x)} \)     M1M1A1

\(x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = f(x),{\text{ }}x > 0\)     AG

Note:     M1 for use of product rule, M1 for use of the fundamental theorem of calculus, A1 for all correct.

METHOD 2

\(x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = f(x)\)

\(\frac{{{\text{d}}(xy)}}{{{\text{d}}x}} = f(x)\)     (M1)

\(xy = \int {f(x){\text{d}}x} \)     M1A1

\(y = \frac{1}{x}\int {f(x){\text{d}}x} \)     AG

[3 marks]

\(y = \frac{1}{x}\left( {2{x^{\frac{1}{2}}} + c} \right)\)     A1A1

Note:     A1 for correct expression apart from the constant, A1 for including the constant in the correct position.

attempt to use the boundary condition     M1

\(c = 4\)     A1

\(y = \frac{1}{x}\left( {2{x^{\frac{1}{2}}} + 4} \right)\)     A1

Note:     Condone use of integrating factor.

[5 marks]

Total [8 marks]

This question allowed for several different approaches. The most common of these was the use of the integrating factor (even though that just took you in a circle). Other candidates substituted the solution into the differential equation and others multiplied the solution by \(x\) and then used the product rule to obtain the differential equation. All these were acceptable.

This was a straightforward question. Some candidates failed to use the hint of ‘hence’, and worked from the beginning using the integrating factor. A surprising number made basic algebra errors such as putting the \( + c\) term in the wrong place and so not dividing it by \(\chi \).