# Solve ( 1 + X ) 2 D 2 Y D X 2 + ( 1 + X ) D Y D X + Y = 4 Cos ( Log ( 1 + X ) ) - Applied Mathematics 2

Sum

Solve (1+x)^2(d^2y)/(dx^2)+(1+x)(dy)/(dx)+y=4cos(log(1+x))

#### Solution

(1+x)^2(d^2y)/(dx^2)+(1+x)(dy)/(dx)+y=4cos(log(1+x))

Put x+1 = v => (dv)/(dx)=1

(dy)/(dx)=(dy)/(dv)

The given eqn changes to ,

v^2(d^2y)/(dv^2)+v(dy)/(dv)+y=4coslogv

Now put log v = z ∴v=𝒆𝒛

[𝑫(𝑫−𝟏)+𝑫+𝟏]𝒚=𝟒𝒄𝒐𝒔 𝒛
∴ (𝑫𝟐+𝟏)𝒚=𝟒𝒄𝒐𝒔 𝒛
For complementary solution ,

𝒇(𝑫)=𝟎
∴ (𝑫𝟐+𝟏)=𝟎
Roots are : i,-i
The complementary solution of given diff. eqn is ,

therefore y_c=c_1cosz+c_2sinz
For particular integral ,

y_p=1/(f(D))x=1/(D^2+1)4cosz=4z/2sinz=2zsinz

therefore y_p=2zsinz

The general solution of given diff. eqn is given by,

y_g=y_c+y_p=c_1cosz+c_2sinz+2zsinz
Resubstitute z and v,

y_g=c_1cos[log(x+1)]+c_2sin[log(1+x)]+2log(1+x)sin[log(1+x)]

Concept: Linear Differential Equation with Constant Coefficient‐ Complementary Function
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