I was lucky enough to have amazing teachers for my GCSEs and A-levels, so I know how helpful it can be to have someone reliable to help you through these crucial periods. I hope that I can provide you with this support and that we can work together to get you the grades you want.

I have previously taught maths and music to younger children, and I have tutored one of my peers at AS level for maths.

I really enjoy teaching, and would really like to expand more into GCSE and A-level; this is where my excitement for maths and German really flourished, so I’d love to be able to pass on my enthusiasm to you through our tutoring sessions.

In the summer of 2015 I completed a Common European Framework C1 level German qualification at Stuttgart University, and keep my German up studying by myself, so I am definitely capable of assisting with German queries as well, although this is not my degree subject.

I look forward to meeting you soon in our sessions! :D

I was lucky enough to have amazing teachers for my GCSEs and A-levels, so I know how helpful it can be to have someone reliable to help you through these crucial periods. I hope that I can provide you with this support and that we can work together to get you the grades you want.

I have previously taught maths and music to younger children, and I have tutored one of my peers at AS level for maths.

I really enjoy teaching, and would really like to expand more into GCSE and A-level; this is where my excitement for maths and German really flourished, so I’d love to be able to pass on my enthusiasm to you through our tutoring sessions.

In the summer of 2015 I completed a Common European Framework C1 level German qualification at Stuttgart University, and keep my German up studying by myself, so I am definitely capable of assisting with German queries as well, although this is not my degree subject.

I look forward to meeting you soon in our sessions! :D

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Claire (Parent from Kettering)

October 15 2015

A very good session my daughter found it to be extremely helpful, this is a great start for Rebecca's tutoring. Thank you Ella

Prepositions affect the case of the noun after them, and two-way prepositions can either put this noun into the dative or the accusative, depending on if there is **movement** in the sentence or not.

Examples of two-way prepositions are:

in

auf

unter

vor

neben

zwischen

an

Most of the time, these prepositions put the noun into the dative. If movement is involved, however, they put the noun into the accusative.

Here are two examples of similar sentences using the preposition "in", one of which uses the dative and one of which uses the accusative.

Ich gehe in die Schule.

Die Maus wohnt in der Schule.

Here, "I" am going to school, so I am moving because I am travelling to school, whereas the mouse lives there all the time and isn't moving. Therefore, "school" is in the accusative in the sentence involving "me", and it's in the dative in the sentence with the mouse.

Warning! Watch out for the verb "bleiben", which means "to stay" - this verb makes the sentence act as though there is movement, even though this verb means the exact opposite!

Prepositions affect the case of the noun after them, and two-way prepositions can either put this noun into the dative or the accusative, depending on if there is **movement** in the sentence or not.

Examples of two-way prepositions are:

in

auf

unter

vor

neben

zwischen

an

Most of the time, these prepositions put the noun into the dative. If movement is involved, however, they put the noun into the accusative.

Here are two examples of similar sentences using the preposition "in", one of which uses the dative and one of which uses the accusative.

Ich gehe in die Schule.

Die Maus wohnt in der Schule.

Here, "I" am going to school, so I am moving because I am travelling to school, whereas the mouse lives there all the time and isn't moving. Therefore, "school" is in the accusative in the sentence involving "me", and it's in the dative in the sentence with the mouse.

Warning! Watch out for the verb "bleiben", which means "to stay" - this verb makes the sentence act as though there is movement, even though this verb means the exact opposite!

We can't just integrate cos^2(x) as it is, so we want to change it into another form, which we can easily do using trig identities.

Recall the double angle formula: cos(2x) = cos^2(x) - sin^2(x). We also know the trig identity sin^2(x) + cos^2(x) = 1, so combining these we get the equation cos(2x) = 2cos^2(x) -1.

Now we can rearrange this to give: cos^2(x) = (1+cos(2x))/2.

So we have an equation which gives cos^2(x) in a nicer form which we can easily integrate using the reverse chain rule.

This eventually gives us an answer of x/2 + sin(2x)/4 +c

:)

We can't just integrate cos^2(x) as it is, so we want to change it into another form, which we can easily do using trig identities.

Recall the double angle formula: cos(2x) = cos^2(x) - sin^2(x). We also know the trig identity sin^2(x) + cos^2(x) = 1, so combining these we get the equation cos(2x) = 2cos^2(x) -1.

Now we can rearrange this to give: cos^2(x) = (1+cos(2x))/2.

So we have an equation which gives cos^2(x) in a nicer form which we can easily integrate using the reverse chain rule.

This eventually gives us an answer of x/2 + sin(2x)/4 +c

:)

Say you have a differential equation of the form:

dy/dx + Py = Q , where P and Q are functions of x.

Remember that even if your equation doesn't initially look like this, you may be able to rearrange it into this format!

The **integrating factor** is e^( ∫P dx).

If we multiply the whole equation by this integrating factor (not forgetting the right hand side!) then it will be in the form:

e^( ∫P dx) dy/dx + Pe^( ∫P dx) y = Qe^( ∫P dx)

This is useful to us because now the left hand side looks like something we might obtain from the product rule ( d/dx(uv)=u'v+v'u), with u=y and v=e^( ∫P dx).

So now we can write:

d/dx (ye^( ∫P dx)) = Qe^( ∫P dx)

Integrating both sides gives:

ye^( ∫P dx) = ∫ (Qe^( ∫P dx)) dx

This is the method, but it may be easier to understand with an example.

y = x dy/dx - 3x

This doesn't look like the form we want, but note that after a little rearranging, it can be expressed as:

dy/dx + y/x = 3

So here we have P=1/x, Q=3.

Our integrating factor is e^( ∫1/x dx) = e^(ln x) = x

Now we want to multiply the whole equation by this integrating factor, giving us:

x dy/dx + y = 3x

Noting that this looks like the product rule with u=y, v=x, we write:

d/dx (xy) = 3x

Integrating both sides gives

xy = ∫3x dx

xy = 1.5x^2 +c

Usually, you'll want the answer to be in the form y= something (although not always), so we divide through by x to give our final answer of:

y = 1.5x +c/x

Say you have a differential equation of the form:

dy/dx + Py = Q , where P and Q are functions of x.

Remember that even if your equation doesn't initially look like this, you may be able to rearrange it into this format!

The **integrating factor** is e^( ∫P dx).

If we multiply the whole equation by this integrating factor (not forgetting the right hand side!) then it will be in the form:

e^( ∫P dx) dy/dx + Pe^( ∫P dx) y = Qe^( ∫P dx)

This is useful to us because now the left hand side looks like something we might obtain from the product rule ( d/dx(uv)=u'v+v'u), with u=y and v=e^( ∫P dx).

So now we can write:

d/dx (ye^( ∫P dx)) = Qe^( ∫P dx)

Integrating both sides gives:

ye^( ∫P dx) = ∫ (Qe^( ∫P dx)) dx

This is the method, but it may be easier to understand with an example.

y = x dy/dx - 3x

This doesn't look like the form we want, but note that after a little rearranging, it can be expressed as:

dy/dx + y/x = 3

So here we have P=1/x, Q=3.

Our integrating factor is e^( ∫1/x dx) = e^(ln x) = x

Now we want to multiply the whole equation by this integrating factor, giving us:

x dy/dx + y = 3x

Noting that this looks like the product rule with u=y, v=x, we write:

d/dx (xy) = 3x

Integrating both sides gives

xy = ∫3x dx

xy = 1.5x^2 +c

Usually, you'll want the answer to be in the form y= something (although not always), so we divide through by x to give our final answer of:

y = 1.5x +c/x