Solve the 2-nd order differential equations y"+2y'=3x

Hello everyone, I'm a new student who just know about Matlab, this is my first exercise, so I wonder if somebody can help me solve this problem.

Solve the 2-nd order differential equations y"+2y'=3x

Thanks a lot, and I will improve it immediately.

Answer 1

[Star Strider]

One approach would be to do this symbolically and solve it using the dsolve function.

[Khang Ng]

I have typed like this:

syms y(x)

Dy= diff(y);

Ode= diff(y,x,2)==-2*diff(y,x)+3*x;

ysol(x)=dsolve(Ode);

ysol=simplify(ysol)

ysol(x) =

C1 - (3*x)/4 + (3*x^2)/4 + C2*exp(-2*x) + 3/8

Can you explain me why we have +3/8 here( while my manual one doesn't have), and does it matter my matlab result, if yes, please tell me how to fix it. Thank you.

[Star Strider]

That may be a constant of integration.

Not entirely trusing my own derivation, I checked with Wolfram Alpha that solved it as:

The Symbolic Math Toolbox solution:

syms y(x)
Dy= diff(y);
Ode= diff(y,x,2)==-2*Dy+3*x;
ysol(x)=dsolve(Ode, Dy(0)==0, y(0)==0);
ysol=simplify(ysol, 500)

ysol(x) = 

figure
fplot(ysol,[0 1])

Nothing that I can find in the documentation gives any clue as to exactly how the was arrived at, and I cannot find any other examples similar to it.

[Khang Ngo]

Oh thank you so much

[Star Strider]

As always, my pleasure!

[Paul]

I'm a bit confused by this thread. The original code at the top of this Answer solved a differential equation with unspecified initial conditions. A subsequent response solved the differential equation assuming zero initial conditions. And there seems to still be an open question about the 3/8.

Is the question just why Matlab chooses not to combine the 3/8 with the C1 in the original code?

[Star Strider]

Setting the initial conditions to 0 was simply to plot the result.

[Paul]

Got it. Is there still an open question about the 3/8?

[Star Strider]

No longer for me. It’s an offset required to satisfy the initial conditions. MATLAB creates (and ) as initial conditions, and with initial conditions of 0 (where this is easiest to see), is 0, since it would then cancel the second term in , that being , satisfying the intital condition.

Wolfram alpha likely includes it in .