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Calculus 1 : Differential Equations

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Example Questions

Example Question #151 : Differential Equations

Find the derivative of the function y=3x^2+2 .

Possible Answers:

y'(x)= 6x+2

y'(x)= 3x^2

y'(x)= 6x

y'(x)= 3x+2

Correct answer:

y'(x)= 6x

Explanation:

To find the derivative of the function, you must use the power rule

$\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ .

Using this rule, the correct answer

$y'=2\cdot 3x^{2-1}$

y'(x)= 6x is obtained.

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Example Question #2472 : Calculus

Find the derivative of the function y=2x^2+5x^4+sin(x) .

Possible Answers:

y'(x)=4x+20x^3+cos(x)

y'(x)=4x+20x^3-sin(x)

y'(x)=4x+20x^3-cos(x)

y'(x)=2x+20x^4-cos(x)

Correct answer:

y'(x)=4x+20x^3+cos(x)

Explanation:

To take the derivative of the first two terms, you use the power rule

$\frac{\mathrm{d} }{\mathrm{d} x}(x^n)=nx^{n-1}$ .

To take the derivative of sin(x), there is the identity that,

sin'(x)=cos(x) .

Using all of the rules and applying,

$y'=2\cdot2x^{2-1}+4\cdot5x^{4-1}+cos(x)$

y'(x)=4x+20x^3+cos(x) .

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Example Question #141 : Solutions To Differential Equations

Determine the solution to the following: y'=15x^2y+12xy+y with y(0)=e^4 .

Possible Answers:

$y=\exp(5x^3+6x^2+x)$

The equation is not separable.

$y=\exp(5x^3+6x^2+x+c)$

$y=\exp(5x^3+6x^2+x+4)$

$y=\exp(5x^3+6x^2+x)+4$

Correct answer:

$y=\exp(5x^3+6x^2+x+4)$

Explanation:

Solving a differential equation tends to be quite simple if it is separable. A separable equation is one that can be rewritten in the following form

$\frac{dy}{dx}= f(x)g(y)$ .

From here we can manipulate the equation

$\frac{1}{g(y)}dy=f(x)dx$

and then you can integrate both sides.

We can rewrite the equation as

$\frac{dy}{dx}=(15x^2+12x+1)y$

and we therefore see that it is separable.

Next we manipulate it to get the the y terms on the left hand side and the x terms on the right hand side like the following

$\frac{dy}{y}=(15x^2+12x+1)dx$ .

The next step is to integrate both sides

$\int\frac{dy}{y}=\int(15x^2+12x+1)dx$ .

This gives us

$\ln y +c_1= 5x^3+6x^2+x+c_2$ ,

since the c's are just arbitrary constants we can combine them to get

$\ln y= 5x^3+6x^2+x+c$ .

Since we want to write this as an equation in terms of y we will apply e to both sides.

Therefore we get

$y=\exp(5x^3+6x^2+x+c)$ .

Since the problem gives us an initial condition we solve for c.

Therefore

$\exp(4)=\exp(5\cdot 0^3+6\cdot 0^2 +0+c)$

so c=4.

Thus the solution is

$y=\exp(5x^3+6x^2+x+4)$ .

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Example Question #144 : How To Find Solutions To Differential Equations

Solve the following differentiable equation:

$\frac{dy}{dx}= \frac{x+y}{x}$

with a condition of y(1)=0 .

Possible Answers:

y=x^2-x-1

y=x+1

y=x^2

y=x-1

y=x^2-x

Correct answer:

y=x^2-x

Explanation:

When given a first order differential equation to solve, one of the first things to check is that it is homogenous.Homogenous equations are ones in which, given

$\frac{dy}{dx}=f(x,y)$ ,

f(x,y)=f(tx,ty) given any t.

If an equation is homogenous you can do a substitution of y=xz and this new equation will be separable.

The first step is to check if the following equation is homogenous,

$\frac{dy}{dx}= \frac{x+y}{x}$ .

Next we show that the equation is homogenous

$f(x,y)=\frac{x+y}{xy}=\frac{t(x+y)}{tx}=\frac{tx+ty}{tx}=f(tx,ty)$ .

So we now replace y with xz. This gives us the following new equation:

$x\frac{dz}{dx}=\frac{x+xz}{x}= 1+z$ .

This is a separable equation which we manipulate and integrate both sides

$\int\frac{1}{x}dx=\int \frac{1}{z+1}dz$

which leads to

$\ln x = \ln(z+1)+c$ .

Thus we get the equation

x=z+1+c ,

then plugging y back in we get

$x=\frac{y}{x}+1+c$ .

So our family of equations is

y=x^2-x-cx

yet plugging in the given condition gives us that c=0.

Thus the final solution

y=x^2-x .

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Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #151 : Differential Equations

Example Question #2472 : Calculus

Example Question #141 : Solutions To Differential Equations

Example Question #144 : How To Find Solutions To Differential Equations

All Calculus 1 Resources

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