2. Problem 3.4.9: y”

Name: 2. Problem 3.4.9: y” - y' = -3
Uploaded: 2022-01-23T08:48:16-08:00
Duration: 7 min 31 s
Channel: Nadir Musofer
Description: 2. Problem 3.4.9: y” - y' = -3

Step 1: The given differential equation is $y''-y'=-3$.

Transcript

00:01 Dear students here we have a question and we have to ask to solve this initial value problem with the initial conditions y of 2 is equal to 3 and y prime of 2 is equal to minus 4 so dear students so dear students here we can write it as here we can write it as d square y divided by d t squared plus y is equal to 0 so here is de students if we put d square y plus y is equal to zero here dearest we have put d is equal to what d divided by d t so here so here this will be this will be de students if we take uh y as a common so the remaining term will be d square plus one two y is equal to zero so these students now uh we know that this is a this is a de student this is a homogeneous homogeneous differential equation so we we will use we have to find the complementary function so for complementary function we need an auxiliary equation to find the roots so we will find the auxiliary equation so the auxiliary equation for this differential equation will be m square plus 1 is equal to 0 so d s students here this is the auxiliary equation which we have so now m square will be equal to minus 1 so the roots will be d s students the roots will be complex so the complex roots are m is equal to plus minus iota so now d s students the complementary function for this will be equal to y is will be equal to d s students the complementary function y is equals to it will be c1 c1 cause x or you can say if we write in a term of t cause t and d students plus c2 sine t so d students here notice that we have what we have the initial conditions 2 so d a student this is 2 so here it will be so here it will be students here it will be what here it will be we can write it as d s students we can write it as c1 cause into t minus 2 plus c2 goes into it is sign distance here it is sine sine t minus so d students here if you and now we have this is the general solution so d students now we have to find now distance we have to find the value of c1 and c2 to find the exact solution so these students we have the initial conditions which is which are day students which are y of 2 is equal to 3 and y prime of 2 is equal to what it is 4 minus 4 so d students here if we use so d students we have to first what we have to first differentiate this function so the differentiation of this function will be y prime is equal to d students this will be minus so d a student this will be minus c1 this will be minus c1 this will be minus c1 so sign into t minus 2 and here it will be plus plus c2 into this will be sign or we can say differentiation of sign is what it is goes into t minus 2 so here dear students so here de students if we put y if we put y of 2 is equal to 3 in now dea students we will use the condition y of 2 is equal to 3 right here right in this in this general equation so this will give de students so this will give de students y is y of y of 2 de students this will give y of 2 is equal to c1 and this will be cause of 0 plus because t if you put t in place of t 2 this will be 0 so d a students right here this will be equal to c2c2 sign of 0 and we know that d a students and we know that y of 2 is equal to what it is 3 is equal to from the initial condition and y of 0 is 0 is equal to 1 and y of sign of 0 is equal to here cause of 0 is equal to 0 so here it will be c1 so c1 will be this now c1 will be equal to what c1 will be equal to 3 so likewise and likewise d .s.

05:25 Students, we have to put y prime of 2 is equal to minus 4.

05:31 In which equation? this equation.

05:35 So this will give d .s.

05:36 So this will give d .s...

Question

2. Problem 3.4.9: $y'' - y' = -3$

Question

2. Problem 3.4.9: $y'' - y' = -3$

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