12th Math Important Questions 2026 | Differential Equations Important Questions | FBISE Math Paper

Añadido: 2026-05-30

[00:00:00]Bismillah Rahma Rahim. Greetings. I have brought a very important topic of second mathematics. Solve homogeneous differential equations with these initial values. Now give you some initial value because you have to solve some differential equation.

[00:00:10]Now look, you cannot separate this differential equation even if you want to.

[00:00:13]So we cannot apply separable equation on this. For this we have to remember only three steps.

[00:00:18]And with those three steps, any homogeneous differential equation can be solved.

[00:00:22]If you learn those three steps, any kind of question can come before you. You can solve it my child. Now what do you have inside the homogeneous differential equation that this type of equation would have been given here. What is basically to be done from this?

[00:00:34]We have the same concept of integrating everything. It's an easy scene.

[00:00:38]Before starting the solution, you just have to remember these three things.

[00:00:40]If you remember these three things, your question will be solved automatically. The first thing you have to remember to solve any differential equation is y = u of x. The second thing that comes out of this is that y / x = u, this is useful for you.

[00:00:56]Ok? Because what do you have to put in place of y/x? u and then its derivative is useful to you.

[00:01:02]Take the derivative of this dy / dx okay? Equals to will apply the product rule to it. The derivative of the first function as it is the second derivative of the second as it is the first. So, if you have any differential equation dy over dx = u + x dudx, then these are the three things we have to remember. One thing, one thing and one thing. If you memorize these three things, you can easily solve any homogeneous differential equation. Now what basically needs to be done? Now you have to isolate dy over dx within this entire equation. Ok? So that we can put the value in its place and make it in the form of y / x.

[00:01:40]So what I have to do is remember these three things. Now let's start with this question. I had solved this but had done it wrong.

[00:01:48]Meaning that I did not write this dx. I said no, don't make any mistake in the beginning. The entire question has to be solved. So what do I have to do, my child?

[00:01:54]Okay, this has to be shifted here. Ok? I teach you step wise. x + y e to the power y x and dx okay? Equ to and x e to the power y x and dy Why did I do this? Because what I have here is I have to draw dy over dx.

[00:02:09]What to make? dy over dx.

[00:02:12]fixed bugs? Now if you look inside this equation, there's y here. What do I want to make? y over x. Just remember that wherever y is, divide x by the entire equation to make it y over x.

[00:02:23]Now what do you want to write here?

[00:02:25]Write here that x divided on both sides. Here we have to write x divide on both sides. If we write x divided on both sides, what will we basically have? x + y e to the power y over x and dx = x e to the power y over x okay? and dy okay? Now inside the power d y and x y and x is made up. Empty we have this y does n't have x. So we divided by x here and also divided by x on the right hand side. The advantage of this is that the question I have will automatically come in the form of y over x.

[00:02:59]So x / x + y / x e to the power y over x bracket close of dx I divided this x by both terms and from here x and x will be subtracted. Just e to the power y over x and dy will come up. fixed bugs? Then this x and this x will also be cut. So 1 + y x e to the power y x and here dx = e to the power y x dy will come. fixed bugs? Then I'll shift dx over there. 1 + y x e to the power y x everywhere pe y over x y x became. So it is beneficial for us that we will put u here. So e to the power y x and dy over dx okay. Now dy over now let's do the substitution. Now we have to put u in place of e y over x and u + x + du in place of dy over dx etc. If we want, we can also divide these e y and x which are multiplied here and gone there.

[00:03:55]But we will do it in the next step after substitution. What do you want to do late? u In whose place will we lie down? What do we do instead of y over x and dy over dx?

[00:04:09]u + x d over dx These two things have to be done. u will come to me 1 + u ee to the power u okay? = u sorry e to the power u na e to the power u okay here instead of u u u and dy and dx we have u + x dud dx if in this substitution we see that if we see whether the trace of y has disappeared or not, the name and trace of y has disappeared this is the best option for us. Now after this we have to separate this equation, separate the separable equation.

[00:04:46]Ok? If we separate it then we will make it separable. fixed bugs? Now to separable, we have the u term all on one side and the x term all on one side. Shift this u here.

[00:04:58]1 + u to the power of e / e to the power of u = okay? This e to the power u multiplied by that got divided. u + x dud dx Now e to the power u divided by both of these. 1 / e to the power u + u e to the power u / e to the power u u + x dud dx e to the power u e to the power u is subtracted what will come 1 / e to the power u + u is just left here and = u + x dud u over dx now this u will be shifted 1 / e to the power u + u - u = x of du over dx u and u is subtracted 1 / e to the power u ^ u = x dudx now after this the work will be easy we will shift it there okay dx will go there so 1 e to the power u of dx x of du and dot will bring it here e u one minute e u is brought here no right now eu is right there it is right there wait a minute x du has to be separated now eu will go there x will come here so 1 / x of dx = e to the power u du now this is my separable equation. It has become separable.

[00:06:20]Now we will do the integrating. So we will do integration.

[00:06:25]Ok? We will integrate both sides.

[00:06:28]fixed bugs? Now I have integration of 1 x of dx = integration of e of ^ u of du. What integration will I get?

[00:06:38]Natural log x = e to the power u so integration of e to the power u is e to the power u plus c. Did you understand till here? Now pick up the calculator. Now we have to solve the initial value problem. Let's look at the initial value problem. e to the power of u + natural log h + c OK? Now we will need the initial value problem.

[00:07:00]How will she be? Look, let's understand a little bit about the initial value problem we have.

[00:07:04]What have you been given? y of 1 = 0 means that the x that we have, that is, we do not read fx = 2, so that means the function itself is two, but what is the image of the function x, I have this value. Now this thing which is basically y of x, I have x. So y of x is = 0 and what do I have? This is y. So if y of 1 = 0.

[00:07:29]It means I have x here?

[00:07:32]x I have one and y I have zero.

[00:07:35]But here y is not there. So we will replace it back. What will you replace? We will replace y and x = u We will replace u. So I have natural log x = e to the power u + c. Let me check a minute. You see e to the power u + c, okay? Ok? Now what will we basically get here? Will do the replacement. Natural log x as it is e to the power u will come in place of e to the power y x + c I will shift it over so natural log x + c do - c do c does not matter okay = e to the power y and x now we will put here what in place of y 0 in place of y and what in place of x? The forest recovered. Now we have natural log 1 + c = e to the power 1 e to the power x has value one and one y is zero so 0 / 1 so natural log 1 + c = e to the power 0 equals one we will calculate natural log 1 on calculator. So the value of natural log 1 is coming out to be zero. So c equals what to me? It became equal to the forest.

[00:08:40]c will be picked up. Will put inside equation number one. natural log x to the power of e, we will not put e to the power of y and x from here because this is my second equation, this is my equation which used y x because y and x were included in the question and plus c so we will put c here, of course put natural log x c here + c = e to the power of y x natural it does not matter, you just put the value inside c + 1 = e to the power of y x so we can consider its solution. e to the power of yx natural log + c what will this leave me with? The solution will be found. fixed bugs? Now the only thing to understand here is that only three equations solve your homogeneous differential equation. What Was the Three Equations? y = ux and y / x = u and y = dy over dx = u + x of dudx These equations will get you solved.

[00:09:43]What will these equations do for you? I will get it solved. Correct? Because what was it about? There was a solution. I hope you guys have understood till here.

[00:09:51]We can solve this question with just these three equations.

[00:09:52]Now send me those questions in the comment box which you find difficult and important, I will solve them here, Insha Allah.

[00:10:01]fixed bugs? So see you in the next video. Please remember me in your prayers. Take care of yourself. Allah Hafiz.