This is a simple example of a first order, separable differential equation. We can move all the \(\raise 0.3pt{y}\) terms to one side and all the \(\raise 0.2pt{x}\) terms to the other. Then we simply integrate both sides with respect to their respective variables.

$$ \begin{eqnarray} 3y\,\small\frac{dy}{dx}\normalsize &=& \small\frac{2x}{y}\normalsize \\[6pt] 3y^2\,dy &=& 2x\,dx \\[6pt] \int 3y^2\,dy &=& \int 2x\,dx \\[6pt] y^3 &=& x^2+c \\[6pt] y &=& \sqrt[\leftroot{-1}\uproot{6}\scriptstyle 3]{x^2+c\:} \\[6pt] \end{eqnarray} $$

Although this looks fairly similar to the first example and is also a first order, separable differential equation, it is significantly more difficult, to the point of being unfair as an exam question.

However, the need to use partial fractions is fair, and it would be great practice if you could follow all these steps and then repeat them for yourself!

$$ \begin{eqnarray} x\,\small\frac{dy}{dx}\normalsize &=& y-y^2 \\[6pt] x\,dy &=& (y-y^2)\,dx \\[6pt] \small\frac{dy}{y-y^2}\normalsize &=& \small\frac{dx}{x}\normalsize \\[6pt] \end{eqnarray} $$

We now need to step away from the solution so that we can express the left-hand integrand in partial fractions.

$$ \small\frac{1}{y-y^2}\normalsize = \small\frac{1}{y(1-y)}\normalsize = \small\frac{A}{y}\normalsize + \small\frac{B}{1-y}\normalsize $$ $$ 1=A(1\!-\!y)+By $$

Substituting \(y\!=\!0\) gives us \(A\!=\!1\) and substituting \(y\!=\!1\) gives us \(B\!=\!1\small.\)

Now we can return to our solution. What makes this so tricky is the need to isolate \(\raise 0.3pt{y}\) so that we can express the solution explicitly.

$$ \begin{eqnarray} \small\frac{dy}{y-y^2}\normalsize &=& \small\frac{dx}{x}\normalsize \\[6pt] \left(\small\frac{1}{y}\normalsize+\small\frac{1}{1-y}\normalsize\right)dy &=& \small\frac{1}{x}\normalsize\,dx \\[6pt] \left(\small\frac{1}{y}\normalsize-\small\frac{1}{y-1}\normalsize\right)dy &=& \small\frac{1}{x}\normalsize\,dx \\[6pt] \int\left(\small\frac{1}{y}\normalsize-\small\frac{1}{y-1}\normalsize\right)dy &=& \int\small\frac{1}{x}\normalsize\,dx \\[6pt] ln\vert y\vert-ln\vert y-1\vert &=& ln\vert x\vert+c \\[6pt] ln\,y-ln\,(y-1) &=& ln\,x+c \\[6pt] ln\,y-ln\,(y-1) &=& ln\,x+ln\,k \\[6pt] ln\,\small\frac{y}{y-1}\normalsize &=& ln\,kx \\[6pt] \small\frac{y}{y-1}\normalsize &=& kx \\[6pt] \small\frac{(y-1)+1}{y-1}\normalsize &=& kx \\[6pt] \small\frac{y-1}{y-1}\normalsize+\small\frac{1}{y-1}\normalsize &=& kx \\[6pt] 1+\small\frac{1}{y-1}\normalsize &=& kx \\[6pt] \small\frac{1}{y-1}\normalsize &=& kx-1 \\[6pt] y-1 &=& \small\frac{1}{kx\!-\!1}\normalsize \\[6pt] \end{eqnarray} $$

$$ \begin{gather} y=\small\frac{1}{kx\!-\!1}\normalsize+1 \\[6pt] y=\small\frac{1}{kx\!-\!1}\normalsize+\small\frac{kx\!-\!1}{kx\!-\!1}\normalsize \\[6pt] y=\small\frac{1+(kx\!-\!1)}{kx\!-\!1}\normalsize \\[6pt] y=\small\frac{kx}{kx\!-\!1}\normalsize \\[6pt] \end{gather} $$

This is also a first order, separable ODE, so we begin as follows:

$$ \begin{gather} \small\frac{dy}{dx}\normalsize = \small\frac{sec\,y}{y}\normalsize \\[6pt] y\ dy = sec\,y\ dx \\[6pt] y\,cos\,y\ dy = 1\ dx \\[6pt] \int y\,cos\,y\ dy = \int 1\ dx \\[6pt] \end{gather} $$

The left hand side is a product of two functions of \(\raise 0.3pt{y}\) so it needs integration by parts.

$$ \begin{matrix} u=y \:&\: v'=cos\,x \\ u'=1 \:&\: v=sin\,x \\ \end{matrix} $$

So we continue like this:

$$ \begin{gather} y\,sin\,y-\int 1.sin\,y\ dy = x+c \\[6pt] y\,sin\,y+cos\,y = x+c \\[6pt] \end{gather} $$

The wording of the question allows us to leave this general solution in implicit form.

Now we substitute the initial conditions \(\raise 0.3pt{x\!=\!\large\frac{\pi}{4}\small,}\) \(\raise 0.3pt{y\!=\!\large\frac{\pi}{2}}\) to work towards the particular solution:

$$ \begin{gather} \small\frac{\pi}{2}\normalsize\,sin\,\small\frac{\pi}{2}\normalsize+cos\,\small\frac{\pi}{2}\normalsize = \small\frac{\pi}{4}\normalsize+c \\[6pt] \small\frac{\pi}{2}\left(1\right)+0 = \small\frac{\pi}{4}\normalsize+c \\[6pt] c = \small\frac{\pi}{2}\normalsize-\small\frac{\pi}{4}\normalsize \\[6pt] c = \small\frac{\pi}{4}\normalsize \\[6pt] \end{gather} $$

So the particular solution is:

$$ y\,sin\,y+cos\,y = x+\small\frac{\pi}{4}\normalsize $$

This isn't separable. It is a relatively simple example of a first order linear differential equation. It is in the format:

$$ \small\frac{dy}{dx}\normalsize+P(x)\small\,\normalsize y=Q(x) $$

In this example:

$$ P(x)=2\small,\normalsize\:Q(x)=5e^{3x} $$

We will show you two methods of solution. Both of them rely on the integrating factor \(I(x)\small,\) defined as follows:

$$ \begin{eqnarray} I(x) &=& e\,^{\int P(x)\,dx} \\[6pt] &=& e\,^{\int 2\,dx} \\[6pt] &=& e^{2x} \\[6pt] \end{eqnarray} $$

Note that we do not need to worry about any constant of integration in the IF.

Method 1: Reverse product rule

Multiply both sides by the IF. The left hand side will take on the format of a product that has been differentiated.

$$ \begin{gather} \small\frac{dy}{dx}\normalsize+2y=5e^{3x} \\[6pt] \underbrace{e^{2x}}_{v}\,\small\underbrace{\frac{dy}{dx}}_{u'}\normalsize+\underbrace{2e^{2x}}_{v'}\underbrace{y}_{u}=5e^{5x} \\[6pt] \small\frac{d}{dx}\normalsize\left(y\small\,\normalsize e^{2x}\right)=5e^{5x} \\[6pt] y\small\,\normalsize e^{2x}=\int 5e^{5x}\,dx \\[6pt] y\small\,\normalsize e^{2x}=e^{5x}+c \\[6pt] y=e^{3x}+\small\frac{c}{e^{2x}} \\[6pt] \end{gather} $$

Method 2: A handy shortcut

The first method is totally logical but a bit laborious. There is a shortcut that can be proven to be an equivalent method, but when first seen it often induces a "Huh?! Where did that come from?" reaction. Anyway, here it is:

$$ I(x)\small\,\normalsize y=\int I(x)\small\,\normalsize Q(x)\,dx $$

If you are happy to take this on trust and learn the shortcut, it can save a little time. Here goes:

$$ \begin{gather} I(x)\small\,\normalsize y=\int I(x)\small\,\normalsize Q(x)\,dx \\[6pt] e^{2x}\small\,\normalsize y=\int e^{2x}.5e^{3x}\,dx \\[6pt] e^{2x}\small\,\normalsize y=\int 5e^{5x}\,dx \\[6pt] e^{2x}\small\,\normalsize y=e^{5x}+c \\[6pt] y=e^{3x}+\small\frac{c}{e^{2x}} \\[6pt] \end{gather} $$

This is another first order linear differential equation. However, to put it into the standard form we first need to divide through by \(\raise 0.2pt{x}\small.\)

$$ \small\frac{dy}{dx}\normalsize+\small\frac{2y}{x}\normalsize=\small\frac{cos\,x}{x}\normalsize $$

Now we can compare this with the standard form of a first order linear ODE:

$$ \small\frac{dy}{dx}\normalsize+P(x)\small\,\normalsize y=Q(x) $$

So we can see that:

$$ P(x)=\small\frac{2}{x}\normalsize\small,\normalsize\:Q(x)=\small\frac{cos\,x}{x}\normalsize $$

Like the previous example, we will show you both methods of obtaining the general solution. You may decide which suits you better. But first, we need the integrating factor:

$$ \begin{eqnarray} I(x) &=& e\,^{\int P(x)\,dx} \\[6pt] &=& e\,^{\int \large\frac{2}{x}\normalsize\,dx} \\[6pt] &=& e\,^{2\,ln\,x} \\[6pt] &=& e\,^{ln\,{x^2}} \\[6pt] &=& x^2 \\[6pt] \end{eqnarray} $$

Method 1: Reverse product rule

Multiply through by the IF and use the product rule backwards. In this example, the right hand side needs integration by parts.

$$ \begin{gather} \small\frac{dy}{dx}\normalsize+\small\frac{2y}{x}\normalsize=\small\frac{cos\,x}{x}\normalsize \\[6pt] x^{2}\small\,\frac{dy}{dx}\normalsize+2xy=x\,cos\,x \\[6pt] \small\frac{d}{dx}\normalsize\left(x^{2}y\right)=x\,cos\,x \\[6pt] x^{2}y=\int x\,cos\,x\ dx \\[6pt] x^{2}y=x\,sin\,x-\int 1.sin\,x\ dx \\[6pt] x^{2}y=x\,sin\,x+cos\,x+c \\[6pt] y=\small\frac{sin\,x}{x}\normalsize+\small\frac{cos\,x}{x^2}\normalsize+\small\frac{c}{x^2}\normalsize \\[6pt] \end{gather} $$

Method 2: The shortcut

This isn't much shorter. Its only slight advantage is that you don't have to think about the product rule. We prefer the first method, but you make up your own mind.

$$ \begin{gather} I(x)\small\,\normalsize y=\int I(x)\small\,\normalsize Q(x)\,dx \\[6pt] x^{2}y=\int x^2.\small\frac{cos\,x}{x}\normalsize\,dx \\[6pt] x^{2}y=\int x\,cos\,x\,dx \\[6pt] x^{2}y=x\,sin\,x-\int 1.sin\,x\ dx \\[6pt] x^{2}y=x\,sin\,x+cos\,x+c \\[6pt] y=\small\frac{sin\,x}{x}\normalsize+\small\frac{cos\,x}{x^2}\normalsize+\small\frac{c}{x^2}\normalsize \\[6pt] \end{gather} $$

This is our first example of a second order linear differential equation.

Because the right hand side is zero we call this homogeneous.

The general form of a homogeneous second order linear differential equation is:

$$ a\small\frac{d^{2}y}{dx^2}\normalsize+b\small\frac{dy}{dx}\normalsize+cy=0 $$

Our first step is to form the auxiliary equation from the left hand side:

$$ \begin{gather} am^2+bm+c=0 \\[6pt] m^2-3m-10=0 \\[6pt] (m+2)(m-5)=0 \\[6pt] m=-2\:\ \small\textsf{or}\normalsize\:\ m=5 \\[6pt] \end{gather} $$

Note that we have two distinct, real roots of the auxiliary equation. That's important. When we have real and distinct roots, \(\raise 0.2pt{p}\) and \(\raise 0.2pt{q}\small,\) say, the general solution of the differential equation is always of the form:

$$ y=Ae^{\tiny\,\normalsize px}+Be^{\tiny\,\normalsize qx} $$

where \(A\) and \(B\) are constants.

So in this example, the general solution is:

$$ y=Ae^{-2x}+Be^{5x} $$

Now we turn our attention to using the initial conditions \(\raise 0.2pt{x\!=\!0}\small,\) \(\raise 0.3pt{y\!=\!2\small,}\) \(\large\frac{dy}{dx}\normalsize\!\!=\!\!-\!11\) to find the particular solution.

First we need to differentiate the general solution to obtain an expression for \(\large\frac{dy}{dx}\small.\)

$$ \small\frac{dy}{dx}\normalsize=-2Ae^{-2x}+5Be^{5x} $$

Now we substitute the initial conditions into our two equations and solve simultaneously for \(A\) and \(B\small.\)

$$ \begin{eqnarray} 2 &=& A+B\:\: && ①\\[6pt] -11 &=& -2A+5B\:\: && ②\\[12pt] 10 &=& 5A+5B\:\: && \small\textsf{5 x }\normalsize ①=③\\[6pt] 21 &=& 7A\:\: && ③-②\\[6pt] \end{eqnarray} $$

So \(A\!=\!3\) and \(B\!=\!\!-\!1\small.\) This gives the particular solution:

$$ y=3e^{-2x}-e^{5x} $$

Like the previous example, this is a second order linear homogeneous differential equation. The general form is:

$$ a\small\frac{d^{2}y}{dx^2}\normalsize+b\small\frac{dy}{dx}\normalsize+cy=0 $$

So we solve the auxiliary equation:

$$ \begin{gather} am^2+bm+c=0 \\[6pt] 9m^2-12m+4=0 \\[6pt] (3m-2)(3m-2)=0 \\[6pt] m=\small\frac{2}{3}\normalsize \\[6pt] \end{gather} $$

Note that we have real and equal roots of the auxiliary equation. That's important. When we have a repeated real root, \(\raise 0.2pt{p}\) say, the general solution of the differential equation is always of the form:

$$ y=Ae^{\tiny\,\normalsize px}+Bxe^{\tiny\,\normalsize px} $$

where \(A\) and \(B\) are constants.

So in this example, the general solution is:

$$ y=Ae^{2x/3}+Bxe^{2x/3} $$

Note that this may also be expressed in factorised form:

$$ y=(A\!+\!Bx)e^{2x/3} $$

Like Examples 6 and 7, this is a second order linear homogeneous differential equation. The general form is:

$$ a\small\frac{d^{2}y}{dx^2}\normalsize+b\small\frac{dy}{dx}\normalsize+cy=0 $$

So we solve the auxiliary equation:

$$ \begin{gather} am^2+bm+c=0 \\[6pt] m^2+2m+5=0 \\[6pt] m=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\[6pt] m=\frac{-2\pm\sqrt{4-20}}{2} \\[6pt] m=\frac{-2\pm\sqrt{-16}}{2} \\[6pt] m=\frac{-2\pm4i}{2} \\[6pt] m=-1\pm 2i \\[6pt] \end{gather} $$

Note that we have complex conjugate roots of the auxiliary equation. That's important. When we have complex conjugate roots \(\raise 0.2pt{p\pm qi\small,}\) the general solution of the differential equation is always of the form:

$$ y=e^{\tiny\,\normalsize px}\left(A\,sin\,qx+B\,cos\,qx\right) $$

where \(A\) and \(B\) are constants.

In this example, \(p\!=\!-\!1\) and \(q\!=\!2\) so the general solution is:

$$ y=e^{-x}\left(A\,sin\,2x+B\,cos\,2x\right) $$

Like the previous three examples, this is a second order linear differential equation, but this time the right hand side is not zero, so it is non-homogeneous.

We start by solving the related homogeneous equation. Its general solution is known as the complementary function, and it forms part of the general solution of the original non-homogeneous equation.

The related homogeneous equation is:

$$ \small\frac{d^{2}y}{dx^2}\normalsize-5\small\frac{dy}{dx}\normalsize+4y=0 $$

The auxiliary equation (AE) is:

$$ \begin{gather} m^2-5m+4=0 \\[6pt] (m-1)(m-4)=0 \\[6pt] m=1\:\ \small\textsf{or}\normalsize\:\ m=4 \\[6pt] \end{gather} $$

The AE has distinct, real roots so the complementary function (CF) is:

$$ y=Ae^{x}+Be^{4x} $$

where \(A\) and \(B\) are constants.

Now we focus on the right hand side of the original ODE: \(\raise 0.2pt{4x-1\small.}\) This is a linear expression so a particular solution of the original equation will be in the form \(y=Cx+D\) where \(\raise 0.1pt{C}\) and \(D\) are constants which we can determine. We start by differentiating twice:

$$ \begin{gather} y=Cx+D \\[6pt] \small\frac{dy}{dx}\normalsize =C \\[6pt] \small\frac{d^{2}y}{dx^2}\normalsize =0 \\[6pt] \end{gather} $$

Now we substitute these back into the original differential equation:

$$ \begin{gather} \small\frac{d^{2}y}{dx^2}\normalsize-5\small\frac{dy}{dx}\normalsize+4y=4x-1 \\[6pt] 0-5C+4(Cx+D)=4x-1 \\[6pt] -5C+4Cx+4D=4x-1 \\[6pt] 4Cx+(4D-5C)=4x-1 \\[6pt] \end{gather} $$

Equating the coefficients of \(\raise 0.2pt{x}\):

$$ \begin{gather} 4C=4 \\[6pt] C=1 \\[6pt] \end{gather} $$

Equating the constant coefficients:

$$ \begin{gather} 4D-5C=-1 \\[6pt] 4D-5=-1 \\[6pt] 4D=4 \\[6pt] D=1 \\[6pt] \end{gather} $$

What we call the particular integral (PI) is therefore \(\raise 0.2pt{Cx+D=x+1\small.}\)

Finally, the general solution of the original equation is just the complementary function plus the particular integral:

$$ y=Ae^{x}+Be^{4x}+x+1 $$

Like the previous example, this is a non-homogeneous second order linear differential equation, although this example is going to have an added complication, as you will see below. Deep breath...

We start by solving the related homogeneous equation:

$$ \small\frac{d^{2}y}{dx^2}\normalsize-4\small\frac{dy}{dx}\normalsize+4y=0 $$

The auxiliary equation (AE) is:

$$ \begin{gather} m^2-4m+4=0 \\[6pt] (m-2)(m-2)=0 \\[6pt] m=2\\[6pt] \end{gather} $$

Because the AE has a repeated real root, the complementary function (CF) is:

$$ y=Ae^{2x}+Bxe^{2x} $$

where \(A\) and \(B\) are constants.

Now we consider the right hand side of the original equation: \(\raise 0.2pt{6e^{2x}\small.}\) Because it is an exponential expression we would usually try \(\raise 0.2pt{y=Ce^{2x}}\) (where \(\raise 0.2pt{C}\) is a constant) as the particular integral (PI).

But here is where the complication arises: there is already a term of that form in the CF: the \(\raise 0.2pt{Ae^{2x}\small.}\) When that happens, we multiply by \(\raise 0.2pt{x}\small,\) giving \(y=Cxe^{2x}\small.\) But that's no good either, because there is also a term of that form in the CF: the \(\raise 0.2pt{Bxe^{2x}\small.}\) When this happens, we need to multiply by \(\raise 0.2pt{x}\) again, so now we have \(\raise 0.2pt{y=Cx^{2}e^{2x}\small.}\)

Now we use the product rule to find the first and second derivatives. Check these for yourself; they're fiddly!

$$ \begin{eqnarray} y&=&Cx^{2}e^{2x} \\[6pt] \small\frac{dy}{dx}\normalsize &=&2Cx^{2}e^{2x}+2Cxe^{2x} \\[6pt] \small\frac{d^{2}y}{dx^2}\normalsize &=& 4Cx^{2}e^{2x}+8Cxe^{2x}+2Ce^{2x}\\[6pt] \end{eqnarray} $$

Next we substitute these back into the original differential equation:

$$ \begin{gather} \small\frac{d^{2}y}{dx^2}\normalsize-4\small\frac{dy}{dx}\normalsize+4y=6e^{2x}\\[10pt] \end{gather} $$

$$ \begin{eqnarray} 4Cx^{2}e^{2x}\!+\!8Cxe^{2x}\!+\!2Ce^{2x}\:\:\:\:\:\:\:\:\:\:\:\:&&\: \\[6pt] -8Cx^{2}e^{2x}-8Cxe^{2x}+4Cx^{2}e^{2x}&=&6e^{2x} \\[10pt] \end{eqnarray} $$

Thankfully we don't need to go any further with this beastly equation! Its sole purpose was to let us find the value of \(\raise 0.2pt{C\small.}\)

So, equating the coefficients of \(\raise 0.2pt{e^{2x}\small,}\) we can see that \(\raise 0.2pt{2C=6}\) so \(\raise 0.2pt{C=3}\) and the particular integral (PI) is therefore \(\raise 0.2pt{3x^{2}e^{2x}\small.}\)

Finally, the general solution of the original equation is just the complementary function plus the particular integral:

$$ y=Ae^{2x}+Bxe^{2x}+3x^{2}e^{2x} $$

This 10-mark question appeared on the SQA 2017 paper .

You may also want to watch a video solution  from DLB Maths.

Like the previous two examples, this is a non-homogeneous second order linear differential equation, although in this example we need to go further and provide a particular solution, not just the general solution.

We start by solving the related homogeneous ODE:

$$ \small\frac{d^{2}y}{dx^2}\normalsize-6\small\frac{dy}{dx}\normalsize+9y=0 $$

The auxiliary equation (AE) is:

$$ \begin{gather} m^2-6m+9=0 \\[6pt] (m-3)^2=0 \\[6pt] m=3\\[6pt] \end{gather} $$

Because the AE has a repeated real root, the complementary function (CF) is:

$$ y=Ae^{3x}+Bxe^{3x} $$

where \(A\) and \(B\) are constants.

Now we consider the right hand side of the original equation: \(8\,sin\,x+19\,cos\,x.\)

Because it is a trigonometic expression we will use \(C\,sin\,x+D\,cos\,x\) (where \(\raise 0.2pt{C}\) and \(\raise 0.2pt{D}\) are constants) as the particular integral.

So, differentiating twice:

$$ \begin{eqnarray} y&=&C\,sin\,x+D\,cos\,x \\[6pt] \small\frac{dy}{dx}\normalsize &=& C\,cos\,x-D\,sin\,x \\[6pt] \small\frac{d^{2}y}{dx^2}\normalsize &=& -\!C\,sin\,x-D\,cos\,x \\[6pt] \end{eqnarray} $$

Next we substitute these back into the original differential equation:

$$ \begin{gather} \small\frac{d^{2}y}{dx^2}\normalsize-6\small\frac{dy}{dx}\normalsize+9y=8\,sin\,x+19\,cos\,x\\[10pt] \end{gather} $$

For clarity, I will split the substitution over several lines:

$$ \begin{gather} -C\,sin\,x-D\,cos\,x \\[6pt] -6(C\,cos\,x-D\,sin\,x) \\[6pt] +9(C\,sin\,x+D\,cos\,x) \\[6pt] \:= 8\,sin\,x+19\,cos\,x \\[10pt] \end{gather} $$

Equating the coefficients of \(sin\,x\):

$$ \begin{eqnarray} -C+6D+9C &=& 8 \\[6pt] 8C+6D &=& 8 \\[6pt] 4C+3D &=& 4\:\:① \\[6pt] \end{eqnarray} $$

Equating the coefficients of \(cos\,x\):

$$ \begin{eqnarray} -D-6C+9D &=& 19 \\[6pt] -6C+8D &=& 19\:\:② \\[6pt] \end{eqnarray} $$

Now we are ready to solve equations ① and ② simultaneously:

$$ \begin{eqnarray} 3\times①:&&\:\:\:12C &+& 9D &=& 12 && \:\:③\\[6pt] 2\times②:&&-\!12C &+& 16D &=& 38 && \:\:④\\[6pt] \end{eqnarray} $$

Eliminating \(\raise 0.2pt{C}\):

$$ \begin{eqnarray} ③+④:\:\:25D &=& 50\\[6pt] D &=& 2\\[6pt] \end{eqnarray} $$

Substituting into ① to find \(\raise 0.2pt{C}\):

$$ \begin{eqnarray} 4C+3D &=& 4\\[6pt] 4C+6 &=& 4\\[6pt] 4C &=& -\!2\\[6pt] C &=& -\!\small\frac12\normalsize \\[6pt] \end{eqnarray} $$

So, adding the CF and PI, we have the following:

$$ y=Ae^{3x}+Bxe^{3x}-\small\frac12\normalsize\,sin\,x+2\,cos\,x $$

Finally, we need to use the given initial conditions to find \(\raise 0.2pt{A}\) and \(\raise 0.2pt{B}\).

Substituting \(y\!=\!7\) and \(x\!=\!0\):

$$ \begin{gather} 7 =Ae^{0}+B(0)e^{0}-\small\frac12\normalsize\,sin\,0+2\,cos\,0\\[6pt] 7 = A+0-0+2\\[6pt] A = 5\\[6pt] \end{gather} $$

To find \(\raise 0.2pt{B,}\) we need to differentiate and then substitute \(\large\frac{dy}{dx}\normalsize\!=\!\large\frac12\normalsize\) and \(x\!=\!0\):

$$ \small\frac{dy}{dx}\normalsize\!=\!3Ae^{3x}\!+\!Be^{3x}(3x\!+\!1)\!-\!\small\frac12\normalsize cos\,x\!-\!2sin\,x $$

$$ \small\frac12\normalsize=15e^0\!+\!Be^{0}(0+1) - \small\frac12\normalsize (1)-2(0) $$

$$ \small\frac12\normalsize=15+B-\small\frac12\normalsize $$

$$ 1=15+B $$

$$ B=-14 $$

Putting this all together, we have the particular solution:

$$ y=5e^{3x}-14xe^{3x}-\small\frac12\normalsize\,sin\,x+2\,cos\,x $$

And that's the end of the last example! If you have studied this far and have been able to reproduce all the examples for yourself, you are most of the way towards being ready for any differential equation that might appear in your Advanced Higher exam.

However, we haven't featured any contextual problems. You will need to study those from past papers and elsewhere. Good luck!