How to solve this first order linear differential equation?

Question

#xy'-1/(x+1)y=x #
y(1) = 0

(According to our professor, I.F. = #e^(intf(x))#, and we should just leave the integral be for now if the integral cannot be solved by hand or conventional methods)

Christopher Agresta · Accepted Answer

# y = x/(x+1)(x + lnx -1) #

We have: # xy'-1/(x+1)y=x # with # y(1) = 0# We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form; # dy/dx + P(x)y=Q(x) # So, we can put the equation in standard form: # y'-1/(x(x+1))y = 1 # Then the integrating factor is given by; # I = e^(int P(x) dx) # # \ \ = exp(int \ -1/(x(x+1)) \ dx) # We can readly evaluate this integral if we perform a partial fraction decomposition of the integrand: # 1/(x(x+1)) -= A/x + B/(x+1) => 1 -= A(x+1)+Bx # Then: # x= \ \ \ \ \ 0 => A=1 # # x=-1 => B=-1# So we can write: # I = exp(int \ -1/x + 1/(x+1) ) \ dx) # # \ \ = exp(ln(x+1)-lnx) # # \ \ = exp( ln((x+1)/x) ) # # \ \ = (x+1)/x # And if we multiply the DE [A] by this Integrating Factor, #I#, we will have a perfect product differential (in fact the original equation); # :. ((x+1)/x)y' - ((x+1)/x) 1/(x(x+1))y = ((x+1)/x) # # :. (1+1/x)y' - 1/x^2y = 1+1/x # # :. d/dx ((1+1/x)y) = 1+1/x # This is now separable, so by "separating the variables" we get: # (1+1/x)y = int \ 1+1/x \ dx # Which is trivial to integrate to get the General Solution: # (1+1/x)y = x + lnx + C # Applying the initial condition #y(1)=0# we get: # 0 = 1 +ln1 + C => C=-1 # Leading to the Particular Solution: # (x+1)/x \ y = x + lnx -1 # # :. y = x/(x+1)(x + lnx -1) #

Paisley Johnson · Answer

#y=x/(x+1)(x+lnx-1)#

.

#xy'-1/(x+1)y=x#

A first Order linear Differential Equation has the form of:

#y'(x)+p(x)y=q(x)#

Let's put our ODE in this form by dividing the entire equation by #x#:

#y'-1/(x(x+1))y=1#, which shows that:

#p(x)=-1/(x(x+1))# and #q(x)=1#

The integration factor is:

#mu(x)=e^(intp(x)dx)=e^(int-1/(x(x+1))dx)=e^I#

#I=int-1/(x(x+1))dx=-int1/(x(x+1))dx#

We can use partial fraction expansion to solve it:

#1/(x(x+1))=A/x+B/(x+1)=(A(x+1)+Bx)/(x(x+1))=(Ax+A+Bx)/(x(x+1))=((A+B)x+A)/(x(x+1))#

#A+B=0# and #A=1#

Therefore, #B=-1#

#I=-int(1/x-1/(x+1))dx=-lnabsx+lnabs(x+1)+C#

#mu(x)=e^(-lnabsx+ln(x+1)+C)=e^-lnabsx*e^(lnabs(x+1))*e^C=1/e^lnabsx(e^lnabs(x+1))e^C=1/x(x+1)e^C=(e^C(x+1))/x#

The constant #e^C# is redundant and can be ignored. Therefore,our integration factor is:

#mu(x)=(x+1)/x#

Now, we multiply both sides of our ODE by this integration factor:

#y'((x+1)/x)-1/(x(x+1))((x+1)/x)y=(x+1)/x#

Then, we simplify and refine:

#((x+1)y')/x-y/x^2=1/x+1# #color(red)(Equation-1)#

If #f=(x+1)/x and g=y# then applying the product rule of differentiation we get:

#(f*g)'=f*g'+f'*g#

#f'=(x-(x+1))/x^2=-1/x^2# (we used the quotient rule)

#g'=y'#

#f*g'+f'*g=((x+1)/x)y'+(-1/x^2)y=((x+1)y')/x-y/x^2#

which is the same as Left Hand Side of #color(red)(Equation-1)#.

Therefore, the Right Hand Side of it must be equal to #(f*g)'#.

#(f*g)'=1/x+1#

#(((x+1)y)/x)'=1/x+1#, i.e:

#d/dx((x+1)/xy)=1/x+1#

We now take the integral of both sides:

#(x+1)/xy=int(1/x+1)dx=int1/xdx+intdx#

#(x+1)/xy=lnx+x+c#

We now proceed to isolate #y#:

Let's multiply both sides by #x#:

#(x+1)y=xlnx+x^2+cx#

Let's divide both sides by #x+1#:

#y=(xlnx+x^2+cx)/(x+1)#

Now, we can apply the initial conditions:

#y(1)=0#

#((1)(ln1)+(1)^2+c(1))/((1)+1)=((1)(0)+1+c)/2=(c+1)/2=0#

#c+1=0#

#c=-1#

Therefore,

#y=(xlnx+x^2-x)/(x+1)=x/(x+1)(x+lnx-1)#

Paisley Johnson · Answer

undefined To solve a first-order linear differential equation, follow these steps:

1. Write the equation in standard form: dy/dx + P(x)y = Q(x), where P(x) and Q(x) are functions of x.

2. Identify the integrating factor, μ(x), defined as e^(∫P(x)dx).

3. Multiply both sides of the equation by the integrating factor, μ(x).

4. Integrate both sides of the equation with respect to x.

5. Solve for y to find the general solution.

6. If an initial condition (boundary condition) is given, use it to find the particular solution by substituting the given values into the general solution.

7. If needed, simplify the solution further.

These steps provide a systematic approach to solve first-order linear differential equations.

How to solve this first order linear differential equation?

#xy'-1/(x+1)y=x # y(1) = 0 (According to our professor, I.F. = #e^(intf(x))#, and we should just leave the integral be for now if the integral cannot be solved by hand or conventional methods)

#xy'-1/(x+1)y=x #
y(1) = 0

(According to our professor, I.F. = #e^(intf(x))#, and we should just leave the integral be for now if the integral cannot be solved by hand or conventional methods)