Question

Q5. (3 marks) Solve the initial value problem (x + 3)y' + y = (x + 3)3, y(0) = 0, and sketch the solution

Answer 1

Given equation is:

We can write it as:

y'+ = (z + 3) (z+3)

Now this is a linear differential equation of first order of the form

In this $P(x)=\frac{1}{x+3} , Q(x)= (x+3)^2$

Integrating factor of this equation is:

$I.F = e^{\int P(x)dx}$

$= e^{\int \frac{1}{(x+3)}dx}$

Solution of this differential equation is:

$y*(I.F) = \int Q(x)*(I.F)dx+ C$

$y(x+3) = \int (x+3)^2*(x+3)dx+ C$

$y(x+3) = \int (x+3)^3dx+ C$

$y(x+3) = \frac{(x+3)^4}{4}+ C$

We can write it as:

$(x+3)y = \frac{(x+3)^4}{4}+ C$

As y(0) = 0

$(0+3)*0 = \frac{(0+3)^4}{4}+ C$

$0 = \frac{81}{4}+ C$

So $C =- \frac{81}{4}$

SO solution of given equation is:

$(x+3)y = \frac{(x+3)^4}{4}-\frac{81}{4}$