# Thread: Differential Equations (3 questions)

1. ## Differential Equations (3 questions)

1 Obtain the General Solution of the differential equations:
(i) $x\frac{dy}{dx} + y = xe^{3x}$

(ii) $\frac{d^2y}{dx^2}=5\frac{dy}{dx}+6y=10e^{-2x}$

2 (a) Find the general solution of the differential equation
$x\frac{dy}{dx}+2y=({1+\frac{2}{x}})e^x$

(b) Find the solution of

$\frac{d^2y}{dx^2}-4\frac{dy}{dx}+5y=2e^{3x}$

given that at x=0,y=2 and $\frac{dy}{dx}=4$

3 (a) Find the following two integrals:
$\int\frac{x}{1-x^2}dx$ and $\int\frac{x}{1-x^2}dx$

Find y in terms of x, givene that
$\frac{dx}{dx}-\frac{x}{1-x^2}y=\frac{x}{(1-x^2)^{\frac{5}{2}}}$

and that y =1 when x = 0.

(b) Solve the differential equation
$\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$

given that y=1 and $\frac{dy}{dx}=3$ when x=0.

2. Originally Posted by Lonehwolf
1 Obtain the General Solution of the differential equations:
(i) $x\frac{dy}{dx} + y = xe^{3x}$

(ii) $\frac{d^2y}{dx^2}=5\frac{dy}{dx}+6y=10e^{-2x}$

2 (a) Find the general solution of the differential equation
$x\frac{dy}{dx}+2y=({1+\frac{2}{x}})e^x$

(b) Find the solution of

$\frac{d^2y}{dx^2}-4\frac{dy}{dx}+5y=2e^{3x}$

given that at x=0,y=2 and $\frac{dy}{dx}=4$

3 (a) Find the following two integrals:
$\int\frac{x}{1-x^2}dx$ and $\int\frac{x}{1-x^2}dx$

Find y in terms of x, givene that
$\frac{dx}{dx}-\frac{x}{1-x^2}y=\frac{x}{(1-x^2)^{\frac{5}{2}}}$

and that y =1 when x = 0.

(b) Solve the differential equation
$\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$

given that y=1 and $\frac{dy}{dx}=3$ when x=0.
1. Notice that the left hand side is the product-rule expansion of

$\frac{d}{dx}(xy)$

So $\frac{d}{dx}(xy) = xe^{3x}$

$xy = \int{xe^{3x}\,dx}$.

You can evaluate the RHS using integration by parts.

3. Originally Posted by Lonehwolf
1 Obtain the General Solution of the differential equations:
(i) $x\frac{dy}{dx} + y = xe^{3x}$
Divide through by the coefficient of $\frac{\,dy}{\,dx}$ to get $\frac{\,dy}{\,dx}+\frac{1}{x}y=e^{3x}$

Apply the technique of the integrating factor here. Can you continue?

(ii) $\frac{d^2y}{dx^2}=5\frac{dy}{dx}+6y=10e^{-2x}$
Do you mean $\frac{\,d^2y}{\,dx^2}{\color{red}-}5\frac{\,dy}{\,dx}+6y=10e^{-2x}$ or $\frac{\,d^2y}{\,dx^2}{\color{red}+}5\frac{\,dy}{\, dx}+6y=10e^{-2x}$??

2 (a) Find the general solution of the differential equation
$x\frac{dy}{dx}+2y=({1+\frac{2}{x}})e^x$
Divide through by the coefficient of $\frac{\,dy}{\,dx}$ to get $\frac{\,dy}{\,dx}+\frac{2}{x}y=\left(\frac{1}{x}+\ frac{2}{x^2}\right)e^x$

Now apply the technique of the integrating factor. Can you continue from here?

(b) Find the solution of

$\frac{d^2y}{dx^2}-4\frac{dy}{dx}+5y=2e^{3x}$

given that at x=0,y=2 and $\frac{dy}{dx}=4$
The characteristic equation is $r^2-4r+5=0$ which means that $r=2\pm i$ (Verify)

Thus, the homogeneous solution is $y_c=e^{2x}\left[c_1\sin\left(x\right)+c_2\cos\left(x\right)\right]$

Now apply the method of undetermined coefficients to determine the particular solution. Once you find the general solution (homogeneous + particular), apply the initial conditions. Can you take it from here?

3 (a) Find the following two integrals:
$\int\frac{x}{1-x^2}dx$ and $\int\frac{x}{1-x^2}dx$
They're the same thing....

Apply the substitution $z=1-x^2$...

Find y in terms of x, givene that
$\frac{dx}{dx}-\frac{x}{1-x^2}y=\frac{x}{(1-x^2)^{\frac{5}{2}}}$

and that y =1 when x = 0.
You should do this with no problem once you have figured out to integrate $\int\frac{x}{1-x^2}\,dx$. Again, you're using the technique of the integrating factor to solve this.

(b) Solve the differential equation
$\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$

given that y=1 and $\frac{dy}{dx}=3$ when x=0.
The corresponding characteristic equation is $r^2-2r+2=0$ which yields $r=1\pm i$ (Verify)

The general solution would be $y=e^x\left[c_1\sin\left(x\right)+c_2\cos\left(x\right)\right]$

Can you take it from here and apply the initial conditions?

Does this make sense?

4. Originally Posted by Lonehwolf
1 Obtain the General Solution of the differential equations:
(i) $x\frac{dy}{dx} + y = xe^{3x}$

(ii) $\frac{d^2y}{dx^2}=5\frac{dy}{dx}+6y=10e^{-2x}$

2 (a) Find the general solution of the differential equation
$x\frac{dy}{dx}+2y=({1+\frac{2}{x}})e^x$

(b) Find the solution of

$\frac{d^2y}{dx^2}-4\frac{dy}{dx}+5y=2e^{3x}$

given that at x=0,y=2 and $\frac{dy}{dx}=4$

3 (a) Find the following two integrals:
$\int\frac{x}{1-x^2}dx$ and $\int\frac{x}{1-x^2}dx$

Find y in terms of x, givene that
$\frac{dx}{dx}-\frac{x}{1-x^2}y=\frac{x}{(1-x^2)^{\frac{5}{2}}}$

and that y =1 when x = 0.

(b) Solve the differential equation
$\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$

given that y=1 and $\frac{dy}{dx}=3$ when x=0.
2. a) This is a first order linear ODE.

So we use the Integrating Factor method.

Divide everything through by x first, to get

$\frac{dy}{dx} + \frac{2}{x}y = \left(\frac{1}{x} + \frac{2}{x^2}\right)e^x$.

Now we find the Integrating Factor and multiply both sides of the equation by it.

It's $e^{\int{\frac{2}{x}\,dx}}=e^{2\ln{x}}=e^{\ln{x^2}} =x^2$.

So the DE becomes

$x^2\frac{dy}{dx} + 2xy = (x + 2)e^x$.

The left hand side is the product rule expansion of $\frac{d}{dx}(x^2y)$

So $\frac{d}{dx}(x^2y) = (x + 2)e^x$

$x^2y = \int{(x + 2)e^x\,dx}$.

Once again, the RHS can be evaluated using Integration by Parts.

5. Thanks for the replies. Regarding the - and + it was - if I remember correctly, and those weren't supposed to be the same thing >.< There was some minor difference.

Alas, I woke up late and couldn't get these done so... i'm as good as roasted, but that's completely my fault for not switching ISPs and for being a blockhead >.>

Thanks for all the inputs, and Pivot thanks for explaining Integrating Factor, I was lost with those for the time being. I may get yelled at but at least I understood 'em xD